Maths-
General
Easy

Question

Which is greater 6 cross times 10 to the power of negative 6 end exponent and 2 cross times 10 to the power of negative 8 end exponent

Hint:

Check whether power of 10 is positive or negative, then move the
decimal places and then compare the values.

The correct answer is: 0.000006


    Complete step by step solution:
    Here 6 cross times 10 to the power of negative 6 end exponent can be written as 0.000006
    Because of the exponent of 10s term is negative, we must move the decimal 6
    places to the left and
    2 cross times 10 to the power of negative 8 end exponent can be written as 0.00000002
    Because of the exponent of 10s term is negative, we must move the decimal 8
    places to the left.
    Here it is clear that 0.000006 > 0.00000002
    That is, 6 cross times 10 to the power of negative 6 end exponent greater than 2 cross times 10 to the power of negative 8 end exponent
     

    Related Questions to study

    General
    Maths-

    Estimate 0.037854921 to the nearest hundredth. Express your answer as a single digit times a power 10 ?
     

    Complete step by step solution:
    Here the digit at the thousandth place in the given decimal number is 7.
    Since 7 > 5,  we add 1 to the digit at the hundredth place and remove all the
    digits right to it.
    So here, we get 0.037854921 rounded to the nearest 100th is 0.04.
    This can be expressed as 4 cross times 10-2

    Estimate 0.037854921 to the nearest hundredth. Express your answer as a single digit times a power 10 ?
     

    Maths-General
    Complete step by step solution:
    Here the digit at the thousandth place in the given decimal number is 7.
    Since 7 > 5,  we add 1 to the digit at the hundredth place and remove all the
    digits right to it.
    So here, we get 0.037854921 rounded to the nearest 100th is 0.04.
    This can be expressed as 4 cross times 10-2
    General
    Maths-

    Sita estimated 30490000000000 as 3 x 108 , What error did she make ?

    Complete step by step solution:
    Here we have  30490000000000 as the number
    We have 10 zeros after 3049,
    hence we can write 30490000000000 as 3049 cross times 10 to the power of 10

    Sita estimated 30490000000000 as 3 x 108 , What error did she make ?

    Maths-General
    Complete step by step solution:
    Here we have  30490000000000 as the number
    We have 10 zeros after 3049,
    hence we can write 30490000000000 as 3049 cross times 10 to the power of 10
    General
    Maths-

    Write the standard form for 6.8 cross times 10 to the power of negative 8 end exponent

    Complete step by step solution:
    Here, 6.8 cross times 10 to the power of negative 8 end exponent equals 0.000000068
    Because of the exponent of 10s term is negative, we must move the decimal 8
    places to the left.

    Write the standard form for 6.8 cross times 10 to the power of negative 8 end exponent

    Maths-General
    Complete step by step solution:
    Here, 6.8 cross times 10 to the power of negative 8 end exponent equals 0.000000068
    Because of the exponent of 10s term is negative, we must move the decimal 8
    places to the left.
    parallel
    General
    Maths-

    Simplify by combining similar terms: 2 root index 8 of 4 plus 7 cube root of 32 minus cube root of 500

    Complete step by step solution:
    Here we can write, 2 cube root of 4 equals 2 cross times 4 to the power of 1 third end exponent comma 7 root index 8 of 32 equals 7 cross times 32 to the power of 1 third end exponent and cube root of 500 equals 500 to the power of 1 third end exponent
    This can be written as 2 cross times 2 to the power of 2 to the power of 1 third end exponent end exponent comma 7 cross times 2 to the power of 5 to the power of 1 third end exponent end exponent text  and  end text open parentheses 2 squared cross times 5 cubed close parentheses to the power of 1 third end exponent
    So, 2 cube root of 4 plus 7 cube root of 32 minus cube root of 500 equals 2 cross times 2 to the power of 2 over 3 end exponent plus 7 cross times 2 to the power of 5 over 3 end exponent minus 5 cross times 2 to the power of 2 over 3 end exponent
    Taking common term 2 to the power of 2 over 3 end exponent outside, we have 2 to the power of 2 over 3 end exponent open parentheses 2 plus 7 cross times 2 to the power of 3 over 3 end exponent minus 5 close parentheses
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 2 to the power of 2 over 3 end exponent open parentheses 2 plus 7 cross times 2 minus 5 close parentheses end cell row cell not stretchy rightwards double arrow 2 to the power of 2 over 3 end exponent left parenthesis 11 right parenthesis end cell end table

    Simplify by combining similar terms: 2 root index 8 of 4 plus 7 cube root of 32 minus cube root of 500

    Maths-General
    Complete step by step solution:
    Here we can write, 2 cube root of 4 equals 2 cross times 4 to the power of 1 third end exponent comma 7 root index 8 of 32 equals 7 cross times 32 to the power of 1 third end exponent and cube root of 500 equals 500 to the power of 1 third end exponent
    This can be written as 2 cross times 2 to the power of 2 to the power of 1 third end exponent end exponent comma 7 cross times 2 to the power of 5 to the power of 1 third end exponent end exponent text  and  end text open parentheses 2 squared cross times 5 cubed close parentheses to the power of 1 third end exponent
    So, 2 cube root of 4 plus 7 cube root of 32 minus cube root of 500 equals 2 cross times 2 to the power of 2 over 3 end exponent plus 7 cross times 2 to the power of 5 over 3 end exponent minus 5 cross times 2 to the power of 2 over 3 end exponent
    Taking common term 2 to the power of 2 over 3 end exponent outside, we have 2 to the power of 2 over 3 end exponent open parentheses 2 plus 7 cross times 2 to the power of 3 over 3 end exponent minus 5 close parentheses
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 2 to the power of 2 over 3 end exponent open parentheses 2 plus 7 cross times 2 minus 5 close parentheses end cell row cell not stretchy rightwards double arrow 2 to the power of 2 over 3 end exponent left parenthesis 11 right parenthesis end cell end table
    General
    Maths-

    Evaluate x squared plus 1 over x squared text  if  end text x equals 3 plus square root of 8

    Complete step by step solution:
    Let x equals 3 plus square root of 8
    ∴ 1 over x equals fraction numerator 1 over denominator 3 plus square root of 8 end fraction equals fraction numerator 1 cross times 3 minus square root of 8 over denominator left parenthesis 3 plus square root of 8 right parenthesis left parenthesis 3 minus square root of 8 right parenthesis end fraction (Rationalising the denominator)
    table attributes columnspacing 1em end attributes row cell equals fraction numerator 3 minus square root of 8 over denominator left parenthesis 3 right parenthesis squared minus left parenthesis square root of 8 right parenthesis squared end fraction equals fraction numerator 3 minus square root of 8 over denominator 9 minus 8 end fraction equals fraction numerator 3 minus square root of 8 over denominator 1 end fraction end cell row cell equals 3 minus square root of 8 end cell end table
    Now, x plus 1 over x equals 3 plus square root of 8 plus 3 minus square root of 8 equals 6
    Squaring both sides, we get open parentheses x plus 1 over x close parentheses squared equals 6 squared
    not stretchy rightwards double arrow x squared plus 1 over x squared plus 2 equals 36
    not stretchy rightwards double arrow x squared plus 1 over x squared equals 36 minus 2 equals 34

    Evaluate x squared plus 1 over x squared text  if  end text x equals 3 plus square root of 8

    Maths-General
    Complete step by step solution:
    Let x equals 3 plus square root of 8
    ∴ 1 over x equals fraction numerator 1 over denominator 3 plus square root of 8 end fraction equals fraction numerator 1 cross times 3 minus square root of 8 over denominator left parenthesis 3 plus square root of 8 right parenthesis left parenthesis 3 minus square root of 8 right parenthesis end fraction (Rationalising the denominator)
    table attributes columnspacing 1em end attributes row cell equals fraction numerator 3 minus square root of 8 over denominator left parenthesis 3 right parenthesis squared minus left parenthesis square root of 8 right parenthesis squared end fraction equals fraction numerator 3 minus square root of 8 over denominator 9 minus 8 end fraction equals fraction numerator 3 minus square root of 8 over denominator 1 end fraction end cell row cell equals 3 minus square root of 8 end cell end table
    Now, x plus 1 over x equals 3 plus square root of 8 plus 3 minus square root of 8 equals 6
    Squaring both sides, we get open parentheses x plus 1 over x close parentheses squared equals 6 squared
    not stretchy rightwards double arrow x squared plus 1 over x squared plus 2 equals 36
    not stretchy rightwards double arrow x squared plus 1 over x squared equals 36 minus 2 equals 34
    General
    Maths-

    Simplify 4√12+5√27 - 3√75 +√300

    Complete step by step solution:
    On prime factorization, we can write
    4 square root of 12 equals 4 cross times square root of 2 cross times 2 cross times 3 end root comma 5 square root of 27 equals 5 cross times square root of 3 cross times 3 cross times 3 end root comma 3 square root of 75 equals 3 cross times square root of 3 cross times 5 cross times 5 end root
    and square root of 300 equals square root of 2 cross times 2 cross times 3 cross times 5 cross times 5 end root
    This can be written as 8 square root of 3 comma 15 square root of 3 comma 15 square root of 3 and 10 square root of 3 respectively.
    So, 4 square root of 12 plus 5 square root of 27 minus 3 square root of 75 plus square root of 300 equals 8 square root of 3 plus 15 square root of 3 minus 15 square root of 3 plus 10 square root of 3 equals 18 square root of 3

    Simplify 4√12+5√27 - 3√75 +√300

    Maths-General
    Complete step by step solution:
    On prime factorization, we can write
    4 square root of 12 equals 4 cross times square root of 2 cross times 2 cross times 3 end root comma 5 square root of 27 equals 5 cross times square root of 3 cross times 3 cross times 3 end root comma 3 square root of 75 equals 3 cross times square root of 3 cross times 5 cross times 5 end root
    and square root of 300 equals square root of 2 cross times 2 cross times 3 cross times 5 cross times 5 end root
    This can be written as 8 square root of 3 comma 15 square root of 3 comma 15 square root of 3 and 10 square root of 3 respectively.
    So, 4 square root of 12 plus 5 square root of 27 minus 3 square root of 75 plus square root of 300 equals 8 square root of 3 plus 15 square root of 3 minus 15 square root of 3 plus 10 square root of 3 equals 18 square root of 3
    parallel
    General
    Maths-

    Simplify √45 - 3√20 +4√5

    Complete step by step solution:
    On prime factorization, we can write
    square root of 45 equals square root of 3 cross times 3 cross times 5 end root comma 3 square root of 20 equals 3 cross times square root of 2 cross times 2 cross times 5 end root text  and  end text 4 square root of 5 equals 4 square root of 5
    This can be written as 3 square root of 5 comma 6 square root of 5 and 4 square root of 5 respectively.
    So, square root of 45 minus 3 square root of 20 plus 4 square root of 5 equals 3 square root of 5 minus 6 square root of 5 plus 4 square root of 5 equals 1 square root of 5 equals square root of 5

    Simplify √45 - 3√20 +4√5

    Maths-General
    Complete step by step solution:
    On prime factorization, we can write
    square root of 45 equals square root of 3 cross times 3 cross times 5 end root comma 3 square root of 20 equals 3 cross times square root of 2 cross times 2 cross times 5 end root text  and  end text 4 square root of 5 equals 4 square root of 5
    This can be written as 3 square root of 5 comma 6 square root of 5 and 4 square root of 5 respectively.
    So, square root of 45 minus 3 square root of 20 plus 4 square root of 5 equals 3 square root of 5 minus 6 square root of 5 plus 4 square root of 5 equals 1 square root of 5 equals square root of 5
    General
    Maths-

    Arrange in ascending order of magnitude square root of 5 comma root index 8 of 11 and 2 root index 6 of 3

    Complete step by step solution:
    Here we can write, square root of 5 equals 5 to the power of 1 half end exponent comma root index 8 of 11 equals 11 to the power of 1 third end exponent text  and  end text 2 root index 6 of 3 equals root index 6 of 12 equals 12 to the power of 1 over 6 end exponent
    On taking the LCM of 2,3,6 we have LCM as 6.
    So, not stretchy rightwards double arrow 5 to the power of 1 half end exponent equals 5 to the power of fraction numerator 1 cross times 3 over denominator 2 cross times 3 end fraction end exponent equals 5 to the power of 3 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 5 cubed close parentheses to the power of 1 over 6 end exponent equals 125 to the power of 1 over 6 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 11 to the power of 1 third end exponent equals 11 to the power of fraction numerator 1 cross times 2 over denominator 3 cross times 2 end fraction end exponent equals 11 to the power of 2 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 11 squared close parentheses to the power of 1 over 6 end exponent equals 121 to the power of 1 over 6 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent
    Now, we have 125 to the power of 1 over 6 end exponent comma 121 to the power of 1 over 6 end exponent and 12 to the power of 1 over 6 end exponent
    So, here ascending order is 12 < 121 < 125
    not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 121 to the power of 1 over 6 end exponent less than 125 to the power of 1 over 6 end exponent
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 11 to the power of 1 third end exponent less than 5 to the power of 1 half end exponent end cell row cell not stretchy rightwards double arrow root index 6 of 12 less than cube root of 11 less than square root of 5 end cell end table

    Arrange in ascending order of magnitude square root of 5 comma root index 8 of 11 and 2 root index 6 of 3

    Maths-General
    Complete step by step solution:
    Here we can write, square root of 5 equals 5 to the power of 1 half end exponent comma root index 8 of 11 equals 11 to the power of 1 third end exponent text  and  end text 2 root index 6 of 3 equals root index 6 of 12 equals 12 to the power of 1 over 6 end exponent
    On taking the LCM of 2,3,6 we have LCM as 6.
    So, not stretchy rightwards double arrow 5 to the power of 1 half end exponent equals 5 to the power of fraction numerator 1 cross times 3 over denominator 2 cross times 3 end fraction end exponent equals 5 to the power of 3 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 5 cubed close parentheses to the power of 1 over 6 end exponent equals 125 to the power of 1 over 6 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 11 to the power of 1 third end exponent equals 11 to the power of fraction numerator 1 cross times 2 over denominator 3 cross times 2 end fraction end exponent equals 11 to the power of 2 over 6 end exponent
    not stretchy rightwards double arrow open parentheses 11 squared close parentheses to the power of 1 over 6 end exponent equals 121 to the power of 1 over 6 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent
    Now, we have 125 to the power of 1 over 6 end exponent comma 121 to the power of 1 over 6 end exponent and 12 to the power of 1 over 6 end exponent
    So, here ascending order is 12 < 121 < 125
    not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 121 to the power of 1 over 6 end exponent less than 125 to the power of 1 over 6 end exponent
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell not stretchy rightwards double arrow 12 to the power of 1 over 6 end exponent less than 11 to the power of 1 third end exponent less than 5 to the power of 1 half end exponent end cell row cell not stretchy rightwards double arrow root index 6 of 12 less than cube root of 11 less than square root of 5 end cell end table
    General
    Maths-

    The Diagonals of a square ABCD meet at O. Prove that A B squared equals 2 A O squared


    Aim  :- Prove that AB squared equals 2 AO squared


    Let length of side of square be a
    Then ,applying using pythagoras theorem in triangle ADC
    We get A C squared equals A D squared plus D C squared not stretchy rightwards double arrow A C squared equals a squared plus a squared
    not stretchy rightwards double arrow AC squared equals 2 straight a squared
    As diagonals bisect each other AC = 2OA
    20 straight A squared equals straight a squared where a =AB
    We get AB squared equals 2 AO squared
    Hence proved

    The Diagonals of a square ABCD meet at O. Prove that A B squared equals 2 A O squared

    Maths-General

    Aim  :- Prove that AB squared equals 2 AO squared


    Let length of side of square be a
    Then ,applying using pythagoras theorem in triangle ADC
    We get A C squared equals A D squared plus D C squared not stretchy rightwards double arrow A C squared equals a squared plus a squared
    not stretchy rightwards double arrow AC squared equals 2 straight a squared
    As diagonals bisect each other AC = 2OA
    20 straight A squared equals straight a squared where a =AB
    We get AB squared equals 2 AO squared
    Hence proved

    parallel
    General
    Maths-

    △ PQR is a right triangle with ∠Q=90 M is the midpoint of QR. Prove that

    P M squared equals P R squared minus 3 Q M squared


    As M is mid point of QR then, QM = ½ RQ
    We get PR squared equals PQ squared plus left parenthesis 2 QM right parenthesis squared not stretchy rightwards double arrow PR squared equals PQ squared plus 4 Q straight Q squared —Eq1
    Applying Pythagoras theorem For triangle PQM
    PM squared equals PQ squared plus QM squared —Eq2
    Substitute Eq2 in Eq1
    We get
    PR squared equals PQ squared plus QM squared plus 3 Q straight Q squared
    not stretchy rightwards double arrow PR squared equals PM squared plus 3 QM squared not stretchy rightwards double arrow PM squared equals PR squared minus 3 QM squared
    Hence proved

    △ PQR is a right triangle with ∠Q=90 M is the midpoint of QR. Prove that

    Maths-General
    P M squared equals P R squared minus 3 Q M squared


    As M is mid point of QR then, QM = ½ RQ
    We get PR squared equals PQ squared plus left parenthesis 2 QM right parenthesis squared not stretchy rightwards double arrow PR squared equals PQ squared plus 4 Q straight Q squared —Eq1
    Applying Pythagoras theorem For triangle PQM
    PM squared equals PQ squared plus QM squared —Eq2
    Substitute Eq2 in Eq1
    We get
    PR squared equals PQ squared plus QM squared plus 3 Q straight Q squared
    not stretchy rightwards double arrow PR squared equals PM squared plus 3 QM squared not stretchy rightwards double arrow PM squared equals PR squared minus 3 QM squared
    Hence proved

    General
    Maths-

    In Triangle A B C comma straight angle A D B equals 90 to the power of ring operator. Prove thatA B squared equals A C squared plus B C squared minus 2 B C times C D

    Aim  :- Prove that  A B squared equals A C squared plus B C squared minus 2 B C times C D
    Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get
    A C squared equals A D squared plus C D squared — Eq1
    Applying pythagoras theorem in Δ ABD ,We get
    A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared minus Eq 2
    Substitute Eq1 in Eq2 ,

    A C squared equals A B squared minus B D squared plus C D squared not stretchy rightwards double arrow A C squared equals A B squared plus C D squared minus B D squared

    text  By using  end text a squared minus b squared equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis text  and  end text B C equals B D plus C D text  we get,  end text

                    not stretchy rightwards double arrow A C squared equals A B squared plus left parenthesis B D plus C D right parenthesis left parenthesis C D minus B D right parenthesis

                   not stretchy rightwards double arrow A C squared equals A B squared plus B C left parenthesis 2 C D minus B C right parenthesis

                      A C squared equals A B squared plus 2 B C. C D minus B C squared

                     A C squared plus B C squared minus 2 B C times C D equals A B squared
    therefore space A B squared equals A C squared plus B C squared minus 2 B C times C D
    Hence proved

    In Triangle A B C comma straight angle A D B equals 90 to the power of ring operator. Prove thatA B squared equals A C squared plus B C squared minus 2 B C times C D

    Maths-General
    Aim  :- Prove that  A B squared equals A C squared plus B C squared minus 2 B C times C D
    Explanation(proof ) :-Applying pythagoras theorem in Δ ACD ,We get
    A C squared equals A D squared plus C D squared — Eq1
    Applying pythagoras theorem in Δ ABD ,We get
    A B squared equals B D squared plus A D squared not stretchy rightwards double arrow A D squared equals A B squared minus B D squared minus Eq 2
    Substitute Eq1 in Eq2 ,

    A C squared equals A B squared minus B D squared plus C D squared not stretchy rightwards double arrow A C squared equals A B squared plus C D squared minus B D squared

    text  By using  end text a squared minus b squared equals left parenthesis a minus b right parenthesis left parenthesis a plus b right parenthesis text  and  end text B C equals B D plus C D text  we get,  end text

                    not stretchy rightwards double arrow A C squared equals A B squared plus left parenthesis B D plus C D right parenthesis left parenthesis C D minus B D right parenthesis

                   not stretchy rightwards double arrow A C squared equals A B squared plus B C left parenthesis 2 C D minus B C right parenthesis

                      A C squared equals A B squared plus 2 B C. C D minus B C squared

                     A C squared plus B C squared minus 2 B C times C D equals A B squared
    therefore space A B squared equals A C squared plus B C squared minus 2 B C times C D
    Hence proved

    General
    Maths-

    In straight triangle P Q R comma Q M perpendicular R P and P R squared minus P Q squared equals Q R squared prove that  QM squared equals PM cross times MR.

    Solution :-
    Aim  :- Prove that Q M squared equals P M cross times M R

    Hint :- using inverse of pythagoras theorem, we get right angle at Q,
    Now using angles prove the similarity of QMR and PMQ triangles. And find the
    ratios to get the desired property
    Explanation(proof ) :-
    Given , P R squared minus P Q squared equals Q R squared not stretchy rightwards double arrow P R squared equals P Q squared plus Q R squared
    Here ,PR is hypotenuse and angle Q is right angle.
    Then straight angle P equals 90 minus straight angle R
    straight angle P equals 90 minus straight angle R text  So we get  end text straight angle P equals straight angle M Q R
    We know straight angle M equals straight angle M equals 90 left parenthesis text  degrees  end text right parenthesis
    By AA similarity

    straight capital delta Q M R tilde straight triangle P M Q

    By similarity we get fraction numerator Q M over denominator P M end fraction equals fraction numerator R M over denominator M Q end fraction not stretchy rightwards double arrow Q M squared equals P M cross times M R
    Hence proved.

     

    In straight triangle P Q R comma Q M perpendicular R P and P R squared minus P Q squared equals Q R squared prove that  QM squared equals PM cross times MR.

    Maths-General
    Solution :-
    Aim  :- Prove that Q M squared equals P M cross times M R

    Hint :- using inverse of pythagoras theorem, we get right angle at Q,
    Now using angles prove the similarity of QMR and PMQ triangles. And find the
    ratios to get the desired property
    Explanation(proof ) :-
    Given , P R squared minus P Q squared equals Q R squared not stretchy rightwards double arrow P R squared equals P Q squared plus Q R squared
    Here ,PR is hypotenuse and angle Q is right angle.
    Then straight angle P equals 90 minus straight angle R
    straight angle P equals 90 minus straight angle R text  So we get  end text straight angle P equals straight angle M Q R
    We know straight angle M equals straight angle M equals 90 left parenthesis text  degrees  end text right parenthesis
    By AA similarity

    straight capital delta Q M R tilde straight triangle P M Q

    By similarity we get fraction numerator Q M over denominator P M end fraction equals fraction numerator R M over denominator M Q end fraction not stretchy rightwards double arrow Q M squared equals P M cross times M R
    Hence proved.

     
    parallel
    General
    Maths-

    In the figure,P Q perpendicular P S comma P Q vertical line vertical line S R comma straight angle S Q R equals 28 to the power of ring operator text  and  end text straight angle Q R T equals 65 to the power of ring operator text ,  end textthen find the value of x and y.

    Explanation :-
    Step 1:-Find ∠QSR
    We  know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.
    ∠QSR +∠SQR = ∠QRT
    ∠QSR +28° = 65°
    ∠QSR = 65° -28° = 37°
    Step 2:- Find value of x
    PQ || SR and QS intersects both the lines
    We get angle x =  ∠QSR ( alternate interior angles)
    x =37°
    Step 3:- Find value of y
    In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)
    90°+37° +y =180 °
    y = 180-127
    y = 53 °

    In the figure,P Q perpendicular P S comma P Q vertical line vertical line S R comma straight angle S Q R equals 28 to the power of ring operator text  and  end text straight angle Q R T equals 65 to the power of ring operator text ,  end textthen find the value of x and y.

    Maths-General
    Explanation :-
    Step 1:-Find ∠QSR
    We  know that in a triangle, the exterior angle is always equal to the sum of the interior opposite angles.
    ∠QSR +∠SQR = ∠QRT
    ∠QSR +28° = 65°
    ∠QSR = 65° -28° = 37°
    Step 2:- Find value of x
    PQ || SR and QS intersects both the lines
    We get angle x =  ∠QSR ( alternate interior angles)
    x =37°
    Step 3:- Find value of y
    In PSQ, ∠P + x + y = 180 °(sum of angles in triangles)
    90°+37° +y =180 °
    y = 180-127
    y = 53 °

    General
    Maths-

    Arrange in descending order of magnitude cube root of 2 comma root index 6 of 3 and root index 9 of 4

    Complete step by step solution:
    Here we can write, root index 8 of 2 equals 2 to the power of 1 third end exponent comma space of 1em root index 6 of 3 equals 3 to the power of 1 over 6 end exponent  and root index 9 of 4 equals 4 to the power of 1 over 9 end exponent
    On taking the LCM of 3,6,9 we have LCM as 18.
    So, not stretchy rightwards double arrow 2 to the power of 1 third end exponent equals 2 to the power of fraction numerator 1 cross times 6 over denominator 3 cross times 6 end fraction end exponent equals 2 to the power of 6 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 2 to the power of 6 close parentheses to the power of 1 over 18 end exponent equals 64 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 3 to the power of 1 over 6 end exponent equals 3 to the power of fraction numerator 1 cross times 3 over denominator 6 cross times 3 end fraction end exponent equals 3 to the power of 3 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 3 cubed close parentheses to the power of 1 over 18 end exponent equals 27 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 4 to the power of 1 over 9 end exponent equals 4 to the power of fraction numerator 1 cross times 2 over denominator 9 cross times 2 end fraction end exponent equals 4 to the power of 2 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 4 squared close parentheses to the power of 1 over 18 end exponent equals 16 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Now, we have 64 to the power of 1 over 18 end exponent comma 27 to the power of 1 over 18 end exponent and 16 to the power of 1 over 18 end exponent
    So, here descending order is 64 > 27 > 16
    not stretchy rightwards double arrow 64 to the power of 1 over 18 end exponent greater than 27 to the power of 1 over 18 end exponent greater than 16 to the power of 1 over 18 end exponent
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 2 to the power of 1 third end exponent greater than 3 to the power of 1 over 6 end exponent greater than 4 to the power of 1 over 9 end exponent end cell row cell not stretchy rightwards double arrow cube root of 2 greater than root index 6 of 3 greater than root index 9 of 4 end cell end table

    Arrange in descending order of magnitude cube root of 2 comma root index 6 of 3 and root index 9 of 4

    Maths-General
    Complete step by step solution:
    Here we can write, root index 8 of 2 equals 2 to the power of 1 third end exponent comma space of 1em root index 6 of 3 equals 3 to the power of 1 over 6 end exponent  and root index 9 of 4 equals 4 to the power of 1 over 9 end exponent
    On taking the LCM of 3,6,9 we have LCM as 18.
    So, not stretchy rightwards double arrow 2 to the power of 1 third end exponent equals 2 to the power of fraction numerator 1 cross times 6 over denominator 3 cross times 6 end fraction end exponent equals 2 to the power of 6 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 2 to the power of 6 close parentheses to the power of 1 over 18 end exponent equals 64 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 3 to the power of 1 over 6 end exponent equals 3 to the power of fraction numerator 1 cross times 3 over denominator 6 cross times 3 end fraction end exponent equals 3 to the power of 3 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 3 cubed close parentheses to the power of 1 over 18 end exponent equals 27 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Likewise, not stretchy rightwards double arrow 4 to the power of 1 over 9 end exponent equals 4 to the power of fraction numerator 1 cross times 2 over denominator 9 cross times 2 end fraction end exponent equals 4 to the power of 2 over 18 end exponent
    not stretchy rightwards double arrow open parentheses 4 squared close parentheses to the power of 1 over 18 end exponent equals 16 to the power of 1 over 18 end exponent space of 1em open parentheses text  Since  end text open parentheses a to the power of b close parentheses to the power of c equals a to the power of b c end exponent close parentheses
    Now, we have 64 to the power of 1 over 18 end exponent comma 27 to the power of 1 over 18 end exponent and 16 to the power of 1 over 18 end exponent
    So, here descending order is 64 > 27 > 16
    not stretchy rightwards double arrow 64 to the power of 1 over 18 end exponent greater than 27 to the power of 1 over 18 end exponent greater than 16 to the power of 1 over 18 end exponent
    table attributes columnspacing 1em end attributes row cell not stretchy rightwards double arrow 2 to the power of 1 third end exponent greater than 3 to the power of 1 over 6 end exponent greater than 4 to the power of 1 over 9 end exponent end cell row cell not stretchy rightwards double arrow cube root of 2 greater than root index 6 of 3 greater than root index 9 of 4 end cell end table
    General
    Maths-

    In △ABC, AD⊥BC Also,B D equals 3 C D Prove that 2 A B squared equals 2 A C squared plus B C squared

    Aim  :- Prove that 2 AB squared equals 2 AC squared plus BC squared
    As BD = 3 CD  ;We get BC = BD +CD = 4 CD
    Applying pythagoras theorem in ΔACD ,We get
    AC squared equals AD squared plus CD squared


    Applying pythagoras theorem in ΔABD ,We get
    AB squared equals BD squared plus AD squared not stretchy rightwards double arrow AD squared equals AB squared minus BD squared  —Eq2
    Substitute Eq2 in Eq1 ,
    AC squared equals AB squared minus BD squared plus CD squared not stretchy rightwards double arrow AC squared equals AB squared plus CD squared minus BD squared
    As BD = 3 CD  ;We get  A C squared equals A B squared plus C D squared minus 9 C D squared
    AC squared equals AB squared minus 8 CD squared not stretchy rightwards double arrow 8 CD squared plus AC squared equals AB squared
    Multiplying with 2 on both sides
    left parenthesis 4 CD right parenthesis squared plus 2 AC squared equals 2 AB squared As BC = 4 CD
    ∴ 2 AB squared equals 2 AC squared plus BC squared
    Hence proved

    In △ABC, AD⊥BC Also,B D equals 3 C D Prove that 2 A B squared equals 2 A C squared plus B C squared

    Maths-General
    Aim  :- Prove that 2 AB squared equals 2 AC squared plus BC squared
    As BD = 3 CD  ;We get BC = BD +CD = 4 CD
    Applying pythagoras theorem in ΔACD ,We get
    AC squared equals AD squared plus CD squared


    Applying pythagoras theorem in ΔABD ,We get
    AB squared equals BD squared plus AD squared not stretchy rightwards double arrow AD squared equals AB squared minus BD squared  —Eq2
    Substitute Eq2 in Eq1 ,
    AC squared equals AB squared minus BD squared plus CD squared not stretchy rightwards double arrow AC squared equals AB squared plus CD squared minus BD squared
    As BD = 3 CD  ;We get  A C squared equals A B squared plus C D squared minus 9 C D squared
    AC squared equals AB squared minus 8 CD squared not stretchy rightwards double arrow 8 CD squared plus AC squared equals AB squared
    Multiplying with 2 on both sides
    left parenthesis 4 CD right parenthesis squared plus 2 AC squared equals 2 AB squared As BC = 4 CD
    ∴ 2 AB squared equals 2 AC squared plus BC squared
    Hence proved

    parallel

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