Maths-
General
Easy

Question

Write the repeating decimal 0.6333....... as a fraction.

Hint:

Perform necessary multiplications to make it as a fraction.

The correct answer is: 19/30


    Complete step by step solution:
    Let x = 0.6333…(i)
    Now multiply by 10 on both the sides in (i),
    We get, 10x = 6.333
    Now multiply by 100 on both the sides in (i),
    100x = 63.33
    On subtracting  from both the sides, we have
    100x - 10x = 63.33 - 6.333
    rightwards double arrow90x = 57
    rightwards double arrowx = 57 over 90
    On simplification, we get 19 over 30
    Hence 0.6333…. = 19 over 30

    Related Questions to study

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    △ ADB and △ CBD are congruent by HL-congruence postulate.

    △ ADB and △ CBD are congruent by HL-congruence postulate.

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    Use the given information to name two triangles that are congruent. Explain your reasoning

    Use the given information to name two triangles that are congruent. Explain your reasoning

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    Use the given information to name two triangles that are congruent. Explain your reasoning. ABCD is a square

    Use the given information to name two triangles that are congruent. Explain your reasoning. ABCD is a square

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    Determine whether each of the following are terminating or repeating
    a) 5.692
    b) 0.22222222222
    c) 7.00001
    d) 7.288888888888
    e) 1.178178178178178.......

    Complete step by step solution:
    a) 5.692 - Terminating decimal
    b) 0.22222222222 - Repeating decimal
    c) 7.00001 - Terminating decimal
    d) 7.288888888888 - Repeating decimal
    e) 1.178178178178178.......- Repeating decimal

    Determine whether each of the following are terminating or repeating
    a) 5.692
    b) 0.22222222222
    c) 7.00001
    d) 7.288888888888
    e) 1.178178178178178.......

    Maths-General
    Complete step by step solution:
    a) 5.692 - Terminating decimal
    b) 0.22222222222 - Repeating decimal
    c) 7.00001 - Terminating decimal
    d) 7.288888888888 - Repeating decimal
    e) 1.178178178178178.......- Repeating decimal
    General
    Maths-

    Name the included angle between the given sides.

    1) AB and AC
    2) AB and BC
    3) AC and CD

    Name the included angle between the given sides.

    1) AB and AC
    2) AB and BC
    3) AC and CD

    Maths-General
    General
    Maths-

    Find the length of the diagonal and area of a Rectangle whose sides are 10 cm and 12 cm. If the perimeter of a square is 2 ( 2x+4y) , find the area ?

    Hint:-
    Area of a rectangle = length × breadth
    All angles in a rectangle are equal to 90°
    Diagonal of a rectangle divides it into 2 right angled triangles.
    Area of a square = side2
    Perimeter of a square = 4 × side
    ep-by-step solution:-
    In the adjacent diagram, we can see that the diagonal divides the given rectangle into 2 right angled triangles.
    Also, diagonal of the rectangle becomes the hypotenuse of the 2 triangles.
    Hypotenuse of the given triangles = diagonal of the given rectangle
    Now, sides of the rectangle = sides of the triangle (other than hypotenuse)
    By applying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴ Hypotenuse2 = 122 + 102
    ∴ Hypotenuse2 = 144 + 100
    ∴ Hypotenuse2 = 244
    ∴ Hypotenuse = 2 √61 cm .............................. (Taking square root both the sides)
    Area of the given rectangle = length × breadth
    ∴ Area of the given rectangle = 12 × 10 ...................................................................................... (From given information & Equation i)
    ∴ Area of the given plot = 120 cm2
    Perimeter of the given square = 4 × sides
    ∴ 2 (2x + 4y) = 4 × sides
    ∴ 4x + 8y = 4 × sides
    ∴ x + 2y = side .................. (Dividing both sides by 4) ................................ (Equation i)
    Now, Area of the given square = side2
    ∴ Area of the given square = (x + 2y)2 ......................................................................................... (From Equation i)
    Final Answer:-
    ∴ Length of the diagonal and area of the given Rectangle is 2 √61 cm & 120 cm2, respectively.
    Area of the given square is (x + 2y)2 cm2.

    Find the length of the diagonal and area of a Rectangle whose sides are 10 cm and 12 cm. If the perimeter of a square is 2 ( 2x+4y) , find the area ?

    Maths-General
    Hint:-
    Area of a rectangle = length × breadth
    All angles in a rectangle are equal to 90°
    Diagonal of a rectangle divides it into 2 right angled triangles.
    Area of a square = side2
    Perimeter of a square = 4 × side
    ep-by-step solution:-
    In the adjacent diagram, we can see that the diagonal divides the given rectangle into 2 right angled triangles.
    Also, diagonal of the rectangle becomes the hypotenuse of the 2 triangles.
    Hypotenuse of the given triangles = diagonal of the given rectangle
    Now, sides of the rectangle = sides of the triangle (other than hypotenuse)
    By applying Pythagorean theorem, For a right angled triangle-
    Hypotenuse2 = sum of the squares of the remaining 2 sides
    ∴ Hypotenuse2 = 122 + 102
    ∴ Hypotenuse2 = 144 + 100
    ∴ Hypotenuse2 = 244
    ∴ Hypotenuse = 2 √61 cm .............................. (Taking square root both the sides)
    Area of the given rectangle = length × breadth
    ∴ Area of the given rectangle = 12 × 10 ...................................................................................... (From given information & Equation i)
    ∴ Area of the given plot = 120 cm2
    Perimeter of the given square = 4 × sides
    ∴ 2 (2x + 4y) = 4 × sides
    ∴ 4x + 8y = 4 × sides
    ∴ x + 2y = side .................. (Dividing both sides by 4) ................................ (Equation i)
    Now, Area of the given square = side2
    ∴ Area of the given square = (x + 2y)2 ......................................................................................... (From Equation i)
    Final Answer:-
    ∴ Length of the diagonal and area of the given Rectangle is 2 √61 cm & 120 cm2, respectively.
    Area of the given square is (x + 2y)2 cm2.
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    Maths-

    Evaluate 8.6 × 10-4 - 4.2 × 10-7

    Hint:-
    In Mathematics, lets say, "a" represents a number and "n" is the number of times that number is to be multiplied with itself, in order to get a desired result. Then, such result can be written as an, where a is called the base and n is its power/ exponent.
    an = a × a × ... × a                      n times.
    Step-by-step solution:-
    8.6 × 10-4 - 4.2 × 10-7
    = 0.00086 - 0.000000042
    = 0.000859958
    = 8.59958 × 10-4
    Final Answer:-
    ∴ Simplifying the given expression, we get- 0.000859958 or 8.59958 × 10-4.

    Evaluate 8.6 × 10-4 - 4.2 × 10-7

    Maths-General
    Hint:-
    In Mathematics, lets say, "a" represents a number and "n" is the number of times that number is to be multiplied with itself, in order to get a desired result. Then, such result can be written as an, where a is called the base and n is its power/ exponent.
    an = a × a × ... × a                      n times.
    Step-by-step solution:-
    8.6 × 10-4 - 4.2 × 10-7
    = 0.00086 - 0.000000042
    = 0.000859958
    = 8.59958 × 10-4
    Final Answer:-
    ∴ Simplifying the given expression, we get- 0.000859958 or 8.59958 × 10-4.
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    Maths-

    A rectangular field has its length and breadth in the ratio 5 : 3. Its area is 3.75 hectares. Find the cost of fencing it at ~ 5 per metre.

    Hint:-
    i. Fencing refers to blocking or covering the boundaries of a land with fences i.e. wooden blocks/ wires to prevent livestock from getting away.
    ii. Perimeter of a rectangle = 2 (length + breadth)
    Step-by-step solution:-
    For the given rectangle field, we are given that length : breadth = 5 : 3
    Let x be the common factor
    ∴ length = 5x, breadth = 3x
    1 hectare = 10,000 sq mt
    ∴ 3.75 hectares = 3.75 × 10,000 = 37,500 m2
    Area of the given rectangular field = length × breadth
    ∴ 37,500 = 5x × 3x ............................................................. (From given information)
    ∴ 37,500 = 15 x2
    ∴ 37,500 / 15 = x2
    ∴ 2,500 = x2
    ∴ 50 = x .................................................................... (Taking square root both the sides)
    ∴ Length = 5x = 5 × 50 = 250 m
    & Breadth = 3x = 3 × 50 = 150 m
    We need to find the cost of fencing the field at Rs. 5 per mt.
    Since fences are put on the boundaries of the field, we need to calculate the perimeter of the field to find the cost of fencing it
    ∴ Perimeter of the given field = 2 (length + breadth)
    ∴ Perimeter of the given field = 2 (250 + 150)
    ∴ Perimeter of the given field = 2 × 400
    ∴ Perimeter of the given field = 800 m
    ∴ cost of fencing the whole field (@ Rs. 5 per mt) = 800 m × Rs. 5/ mt
    ∴ cost of fencing the whole field (@ Rs. 5 per mt) = Rs. 4,000
    Final Answer:-
    ∴ The cost of fencing the given rectangle field is Rs. 4,000.

    A rectangular field has its length and breadth in the ratio 5 : 3. Its area is 3.75 hectares. Find the cost of fencing it at ~ 5 per metre.

    Maths-General
    Hint:-
    i. Fencing refers to blocking or covering the boundaries of a land with fences i.e. wooden blocks/ wires to prevent livestock from getting away.
    ii. Perimeter of a rectangle = 2 (length + breadth)
    Step-by-step solution:-
    For the given rectangle field, we are given that length : breadth = 5 : 3
    Let x be the common factor
    ∴ length = 5x, breadth = 3x
    1 hectare = 10,000 sq mt
    ∴ 3.75 hectares = 3.75 × 10,000 = 37,500 m2
    Area of the given rectangular field = length × breadth
    ∴ 37,500 = 5x × 3x ............................................................. (From given information)
    ∴ 37,500 = 15 x2
    ∴ 37,500 / 15 = x2
    ∴ 2,500 = x2
    ∴ 50 = x .................................................................... (Taking square root both the sides)
    ∴ Length = 5x = 5 × 50 = 250 m
    & Breadth = 3x = 3 × 50 = 150 m
    We need to find the cost of fencing the field at Rs. 5 per mt.
    Since fences are put on the boundaries of the field, we need to calculate the perimeter of the field to find the cost of fencing it
    ∴ Perimeter of the given field = 2 (length + breadth)
    ∴ Perimeter of the given field = 2 (250 + 150)
    ∴ Perimeter of the given field = 2 × 400
    ∴ Perimeter of the given field = 800 m
    ∴ cost of fencing the whole field (@ Rs. 5 per mt) = 800 m × Rs. 5/ mt
    ∴ cost of fencing the whole field (@ Rs. 5 per mt) = Rs. 4,000
    Final Answer:-
    ∴ The cost of fencing the given rectangle field is Rs. 4,000.
    General
    Maths-

    What additional information is necessary to prove that the triangles ABC and XYZ are congruent by HL-Congruence theorem?

    What additional information is necessary to prove that the triangles ABC and XYZ are congruent by HL-Congruence theorem?

    Maths-General
    parallel
    General
    Maths-

    Find the value of ( 50 - 40) x 8-1

    Hint:-
    In Mathematics, lets say, "a" represents a number and "n" is the number of times that number is to be multiplied with itself, in order to get a desired result. Then, such result can be written as an, where a is called the base and n is its power/ exponent.
    an = a × a × ... × a           n times.
    Step-by-step solution:-
    (50 - 40) × 8-1
    = (1-1) × 1/8 …............................... (b0 = 1 & b-n = 1/bn)
    = 0 × 1/8
    = 0
    Final Answer:-
    ∴ Simplifying the given expression, we get- 0.

    Find the value of ( 50 - 40) x 8-1

    Maths-General
    Hint:-
    In Mathematics, lets say, "a" represents a number and "n" is the number of times that number is to be multiplied with itself, in order to get a desired result. Then, such result can be written as an, where a is called the base and n is its power/ exponent.
    an = a × a × ... × a           n times.
    Step-by-step solution:-
    (50 - 40) × 8-1
    = (1-1) × 1/8 …............................... (b0 = 1 & b-n = 1/bn)
    = 0 × 1/8
    = 0
    Final Answer:-
    ∴ Simplifying the given expression, we get- 0.
    General
    Maths-

    Are the given triangles congruent? Explain why or why not.

    Are the given triangles congruent? Explain why or why not.

    Maths-General
    General
    Maths-

    Are the given triangles congruent? Explain why or why not

    Are the given triangles congruent? Explain why or why not

    Maths-General
    parallel
    General
    Maths-

    Find the value of
    (67.542)2 - (32.458)2 / 75.458 - 40.374

    Hint:-
    In Mathematics, lets say, "a" represents a number and "n" is the number of times that number is to be multiplied with itself, in order to get a desired result. Then, such result can be written as an, where a is called the base and n is its power/ exponent.
    an = a × a × ... × a       n times.
    Step-by-step solution:-
    (67.542)2 - (32.458)2 / 75.458 - 40.374
    = 4561.92 - 1053.52 / 35.084
    = 3508.4 / 35.084
    = 100
    Final Answer:-
    ∴ Simplifying the given expression, we get- 100.

    Find the value of
    (67.542)2 - (32.458)2 / 75.458 - 40.374

    Maths-General
    Hint:-
    In Mathematics, lets say, "a" represents a number and "n" is the number of times that number is to be multiplied with itself, in order to get a desired result. Then, such result can be written as an, where a is called the base and n is its power/ exponent.
    an = a × a × ... × a       n times.
    Step-by-step solution:-
    (67.542)2 - (32.458)2 / 75.458 - 40.374
    = 4561.92 - 1053.52 / 35.084
    = 3508.4 / 35.084
    = 100
    Final Answer:-
    ∴ Simplifying the given expression, we get- 100.
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    Maths-

    Why is SSA not a valid method for proving that triangles are congruent?

    Complete step by step solution:
    SSA is not possible since the sides could be located in two different parts of the
    triangles and not the corresponding sides of two triangles. There is chance that the
    size and shape would be different for both triangles and for triangles to be congruent,
    the triangles must be of the same length, shape and size.

    Why is SSA not a valid method for proving that triangles are congruent?

    Maths-General
    Complete step by step solution:
    SSA is not possible since the sides could be located in two different parts of the
    triangles and not the corresponding sides of two triangles. There is chance that the
    size and shape would be different for both triangles and for triangles to be congruent,
    the triangles must be of the same length, shape and size.
    General
    Maths-

    Prove that △ ACB ≅△ CAD by SAS postulate.

    Prove that △ ACB ≅△ CAD by SAS postulate.

    Maths-General
    parallel

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