Maths-
General
Easy

Question

Find the volume of a sphere whose surface area is 154 sq.cm

Hint:

The surface area of a sphere is 4 pi r squared

The correct answer is: 179.67cm3


    Explanation:
    • We have given surface area 154 cm squared.
    • We have to find the volume of the sphere.
    Step 1 of 1:
    We have given surface area 154cm2.
    So,
    4 pi r squared equals 154 cm squared
    r squared equals fraction numerator 154 over denominator 4 pi end fraction
    r squared equals 12.25
    r = 3.5
    Volume of sphere
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 3.5 right parenthesis cubed
    equals 179.67 cm cubed
    Thus, the volume of the sphere is 179.67cm3.

    Related Questions to study

    General
    Maths-

    A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m , then find the volume of the iron used to make the tank ?

    Explanation:
    • We have given a hemisphere with thickness 1cm.
    • We have to find the volume of iron used to make the tank.
    Step 1 of 1:
    Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.
    The Volume of hemisphere of base radius r is equal to 2 over 3 pi r cubed
    The inner radius of the tank r = 1m
    Thickness of iron = 1cm = 1/100 m =0.01m
    Outer radius of the tank, R = 1m + 0.01m = 1.01m
    The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.
    Volume of the iron used to make the tank 2 over 3 pi R cubed minus 2 over 3 pi r cubed
    equals 2 divided by 3 pi open parentheses R cubed minus r squared close parentheses
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets left parenthesis 1.01 m right parenthesis cubed minus left parenthesis 1 m right parenthesis cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets 1.030301 m cubed minus 1 m cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times 0.030301 straight m cubed
    equals 0.06348 straight m cubed text  (approx.)  end text
    0.06348 mof iron used to make the tank

    A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m , then find the volume of the iron used to make the tank ?

    Maths-General
    Explanation:
    • We have given a hemisphere with thickness 1cm.
    • We have to find the volume of iron used to make the tank.
    Step 1 of 1:
    Since the hemispherical tank is made of 1cm thick iron, we can find the outer radius of the tank by adding thickness to the inner radius.
    The Volume of hemisphere of base radius r is equal to 2 over 3 pi r cubed
    The inner radius of the tank r = 1m
    Thickness of iron = 1cm = 1/100 m =0.01m
    Outer radius of the tank, R = 1m + 0.01m = 1.01m
    The volume of the iron used to make the tank can be calculated by subtracting the volume of the tank with inner radius from the volume of the tank with outer radius.
    Volume of the iron used to make the tank 2 over 3 pi R cubed minus 2 over 3 pi r cubed
    equals 2 divided by 3 pi open parentheses R cubed minus r squared close parentheses
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets left parenthesis 1.01 m right parenthesis cubed minus left parenthesis 1 m right parenthesis cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times open square brackets 1.030301 m cubed minus 1 m cubed close square brackets
    equals 2 divided by 3 cross times 22 divided by 7 cross times 0.030301 straight m cubed
    equals 0.06348 straight m cubed text  (approx.)  end text
    0.06348 mof iron used to make the tank
    General
    Maths-

    How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

    Explanation:
    • We have given a hemispherical bowl of diameter 10.5cm
    • We have to find how many litre of milk this bowl can hold.
    Step 1 of 1:
    The diameter of the hemispherical bowl is
    Radius will be fraction numerator 10.5 cm over denominator 2 end fraction equals 5.25 cm
    So, The volume of the hemispherical bowl will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 5.25 right parenthesis cubed
    equals 303.065 cm cubed
    Also we know that 1 cm cubed equals 0.001 text  liter  end text
    So,
    303.065 cm cubed equals 0.3030 text  liter  end text

    How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

    Maths-General
    Explanation:
    • We have given a hemispherical bowl of diameter 10.5cm
    • We have to find how many litre of milk this bowl can hold.
    Step 1 of 1:
    The diameter of the hemispherical bowl is
    Radius will be fraction numerator 10.5 cm over denominator 2 end fraction equals 5.25 cm
    So, The volume of the hemispherical bowl will be
    V equals 2 over 3 pi r cubed
    equals 2 over 3 pi left parenthesis 5.25 right parenthesis cubed
    equals 303.065 cm cubed
    Also we know that 1 cm cubed equals 0.001 text  liter  end text
    So,
    303.065 cm cubed equals 0.3030 text  liter  end text
    General
    Maths-

    straight angle A of an isosceles triangle ABC is acute in which A B equals A C text  and  end text B D perpendicular A C and . Prove that
    B C squared equals 2 cross times A C cross times C D

    Solution :-
    Aim  :- proveB C squared equals 2 ∗ A C ∗ C D
    Hint :- Using pythagoras theorem, find the length of  side AB and BC .
    Substitute the RHS of B C squared equals C D squared plus B D squared in terms of AC and CD Explanation(proof 
    Using pythagoras theorem,in triangle BDC
    B C squared equals C D squared plus B D squared— Eq1
    Using pythagoras theorem,in triangle BDA                                                                                                      A B squared equals B D squared plus A D squared not stretchy rightwards double arrow B D squared equals A B squared minus A D squared— Eq2
    Substitute Eq2 in Eq1
    B C squared equals A B squared minus A D squared plus C D squared
    Substitute AD = AC - CD (from diagram)
    B C squared equals A B squared minus left parenthesis A C minus C D right parenthesis squared plus C D squared not stretchy rightwards double arrow B C squared
    equals A B squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    text  As  end text A B equals A C left parenthesis text  given  end text right parenthesis text  we get  end text comma B C squared equals A C squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    not stretchy rightwards double arrow B C squared equals 2 ∗ A C ∗ C D
    Hence proved

     

    straight angle A of an isosceles triangle ABC is acute in which A B equals A C text  and  end text B D perpendicular A C and . Prove that
    B C squared equals 2 cross times A C cross times C D

    Maths-General
    Solution :-
    Aim  :- proveB C squared equals 2 ∗ A C ∗ C D
    Hint :- Using pythagoras theorem, find the length of  side AB and BC .
    Substitute the RHS of B C squared equals C D squared plus B D squared in terms of AC and CD Explanation(proof 
    Using pythagoras theorem,in triangle BDC
    B C squared equals C D squared plus B D squared— Eq1
    Using pythagoras theorem,in triangle BDA                                                                                                      A B squared equals B D squared plus A D squared not stretchy rightwards double arrow B D squared equals A B squared minus A D squared— Eq2
    Substitute Eq2 in Eq1
    B C squared equals A B squared minus A D squared plus C D squared
    Substitute AD = AC - CD (from diagram)
    B C squared equals A B squared minus left parenthesis A C minus C D right parenthesis squared plus C D squared not stretchy rightwards double arrow B C squared
    equals A B squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    text  As  end text A B equals A C left parenthesis text  given  end text right parenthesis text  we get  end text comma B C squared equals A C squared minus A C squared minus C D squared plus 2 ∗ A C ∗ C D plus C D squared
    not stretchy rightwards double arrow B C squared equals 2 ∗ A C ∗ C D
    Hence proved

     
    parallel
    General
    Maths-

    The Diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

    Explanation:
    • We have given the diameter of moon is approximately one fourth of the diameter of the earth.
    • We have to find the fraction of the volume of the earth is the volume of the moon
    Step 1 of 1:
    Let the diameter of moon be ,Then the diameter of earth will be 4d
    So, Radius of moon will be d over 2. And the radius of earth will be fraction numerator 4 d over denominator 2 end fraction equals 2 d
    Since, both earth and moon are sphere
    We can use formula of volume of sphere 4 over 3 pi r cubed.
    V subscript m equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses d over 2 close parentheses cubed
    equals fraction numerator pi d cubed over denominator 6 end fraction
    And volume of earth will be
    V subscript E equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 2 d right parenthesis cubed
    equals fraction numerator 32 pi d cubed over denominator 3 end fraction
    Therefore, the fraction will be
    V subscript m over V subscript E equals fraction numerator pi d cubed divided by 6 over denominator 32 pi d cubed divided by 3 end fraction
    equals 1 over 64

    The Diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

    Maths-General
    Explanation:
    • We have given the diameter of moon is approximately one fourth of the diameter of the earth.
    • We have to find the fraction of the volume of the earth is the volume of the moon
    Step 1 of 1:
    Let the diameter of moon be ,Then the diameter of earth will be 4d
    So, Radius of moon will be d over 2. And the radius of earth will be fraction numerator 4 d over denominator 2 end fraction equals 2 d
    Since, both earth and moon are sphere
    We can use formula of volume of sphere 4 over 3 pi r cubed.
    V subscript m equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses d over 2 close parentheses cubed
    equals fraction numerator pi d cubed over denominator 6 end fraction
    And volume of earth will be
    V subscript E equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 2 d right parenthesis cubed
    equals fraction numerator 32 pi d cubed over denominator 3 end fraction
    Therefore, the fraction will be
    V subscript m over V subscript E equals fraction numerator pi d cubed divided by 6 over denominator 32 pi d cubed divided by 3 end fraction
    equals 1 over 64
    General
    Maths-

    The diameter of a metallic ball is 4.2 cm. What is the mass of the ball , if the density of the metal is 8.9 g per cubic cm ?

    Explanation:
    • We have given diameter of metallic ball .and the density of the metal is
    • We have to find the mass of the ball.
    Step 1 of 1:
    We will first find the volume of the metallic ball
    V equals 4 over 3 pi r cubed
    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 equals 4 over 3 pi left parenthesis 4.2 right parenthesis cubed end cell row cell equals 310.339 cm cubed end cell end table
    Now we know that
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    Put the value of density and volume in formula, we will get mass
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    8.9 = fraction numerator M a s s over denominator 310.399 end fraction
    Mass = 8.9 cross times 310.399gm
    = 2762.5511gm

    The diameter of a metallic ball is 4.2 cm. What is the mass of the ball , if the density of the metal is 8.9 g per cubic cm ?

    Maths-General
    Explanation:
    • We have given diameter of metallic ball .and the density of the metal is
    • We have to find the mass of the ball.
    Step 1 of 1:
    We will first find the volume of the metallic ball
    V equals 4 over 3 pi r cubed
    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 equals 4 over 3 pi left parenthesis 4.2 right parenthesis cubed end cell row cell equals 310.339 cm cubed end cell end table
    Now we know that
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    Put the value of density and volume in formula, we will get mass
    Density = fraction numerator M a s s over denominator V o l u m e end fraction
    8.9 = fraction numerator M a s s over denominator 310.399 end fraction
    Mass = 8.9 cross times 310.399gm
    = 2762.5511gm
    General
    Maths-

    Find the amount of water displaced by a solid spherical ball of diameter 0.21m

    Explanation:
    • We have given a solid spherical ball with diameter
    • We have to find the volume of water displaced by the given sphere.
    Step 1 of 1:
    We know that the volume of water displaced by the given sphere will be equal to volume of sphere.
    So,
    Volume of sphere will be

    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses fraction numerator 0.21 over denominator 2 end fraction close parentheses cubed
    equals 0.00484 m cubed

    Find the amount of water displaced by a solid spherical ball of diameter 0.21m

    Maths-General
    Explanation:
    • We have given a solid spherical ball with diameter
    • We have to find the volume of water displaced by the given sphere.
    Step 1 of 1:
    We know that the volume of water displaced by the given sphere will be equal to volume of sphere.
    So,
    Volume of sphere will be

    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi open parentheses fraction numerator 0.21 over denominator 2 end fraction close parentheses cubed
    equals 0.00484 m cubed

    parallel
    General
    Maths-

    Write truth table for the converse of p → q.

    Hint:
    If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p not stretchy rightwards arrow q is
    q not stretchy rightwards arrow p
    Solution
    The converse of the conditional statement p not stretchy rightwards arrow q is q not stretchy rightwards arrow  p

    Final Answer:
    Hence, we have drawn the truth table above.

     

    Write truth table for the converse of p → q.

    Maths-General
    Hint:
    If a conditional statement of the form "If p then q" is given. The converse is "If q then p." Symbolically, the converse of p not stretchy rightwards arrow q is
    q not stretchy rightwards arrow p
    Solution
    The converse of the conditional statement p not stretchy rightwards arrow q is q not stretchy rightwards arrow  p

    Final Answer:
    Hence, we have drawn the truth table above.

     
    General
    Maths-

    Find the volume of a sphere whose radius is 0.63 m

    Explanation:
    • We have given a sphere with radius .
    • We have to find the volume of sphere.
    Step 1 of 1:
    We know that the volume of sphere with radius  is .
    Now put the value of
    So,
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 0.63 right parenthesis cubed
    equals 1.0473 straight m cubed

    Find the volume of a sphere whose radius is 0.63 m

    Maths-General
    Explanation:
    • We have given a sphere with radius .
    • We have to find the volume of sphere.
    Step 1 of 1:
    We know that the volume of sphere with radius  is .
    Now put the value of
    So,
    V equals 4 over 3 pi r cubed
    equals 4 over 3 pi left parenthesis 0.63 right parenthesis cubed
    equals 1.0473 straight m cubed
    General
    Maths-

    Make a valid conclusion in the situation.
    If two points lie in a plane, then the line containing them lies in the plane.
    Points A and B lie in plane PQR.

    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Two points lie in a plane
    q: Line containing the two points is in the plane
    So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as
    not stretchy rightwards arrow q
    We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B  lie in the plane PQR.
    Final Answer:
    Hence we can conclude that the line containing points A and B  lie in the plane PQR.

    Make a valid conclusion in the situation.
    If two points lie in a plane, then the line containing them lies in the plane.
    Points A and B lie in plane PQR.

    Maths-General
    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Two points lie in a plane
    q: Line containing the two points is in the plane
    So we can write the given statement “If two points lie in a plane, then the line containing them lies in the plane” as
    not stretchy rightwards arrow q
    We are given that two points A and B lie in plane PQR so we can conclude that the line containing the points A and B  lie in the plane PQR.
    Final Answer:
    Hence we can conclude that the line containing points A and B  lie in the plane PQR.
    parallel
    General
    Maths-

    The conditional p → q is only false when
    a. p = T, q = T
    b. p = T, q = F
    c. p = F, q = F
    d. p = F, q = T

    Hint:
    p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true
    Solution
    The truth table of p → q is given as

    Hence, we can see that p → q is only false when p is True and q is False.
    Final Answer:
    So, p → q is only false when p = T and q = F. Hence option b is correct.

    The conditional p → q is only false when
    a. p = T, q = T
    b. p = T, q = F
    c. p = F, q = F
    d. p = F, q = T

    Maths-General
    Hint:
    p → q is a conditional statement which mean if p then q. The conditional statement is saying that if p is true, then q will immediately follow and thus be true
    Solution
    The truth table of p → q is given as

    Hence, we can see that p → q is only false when p is True and q is False.
    Final Answer:
    So, p → q is only false when p = T and q = F. Hence option b is correct.
    General
    Maths-

    Use p and q to write the symbolic statement in words.
    p: Roses are red.
    q: Roses are beautiful.
    I) p → q
    II)  ~q
    III) ~q → ~p

    Hint:
    p → q represents the conditional statement which means “if p then q” or “p implies q”
    “ ~ ” symbol represents the inverse of a statement and inverse of a conditional statement like   p → q is given as ~p → ~q which means if not p then not q
    Solution
    It is given
    p: Roses are red.
    q: Roses are beautiful.
    Symbolic statement of p → q is “ If Roses are red, then the roses are beautiful. ”
    Symbolic statement of ~q is “ Roses are not beautiful ”
    Symbolic statement of ~q → ~p is “ If Roses are not red, then the roses are not beautiful. ”
    Final Answer:
    p → q : If Roses are red, then the roses are beautiful
    ~q: Roses are not beautiful
    ~q → ~p: If Roses are not red, then the roses are not beautiful.

    Use p and q to write the symbolic statement in words.
    p: Roses are red.
    q: Roses are beautiful.
    I) p → q
    II)  ~q
    III) ~q → ~p

    Maths-General
    Hint:
    p → q represents the conditional statement which means “if p then q” or “p implies q”
    “ ~ ” symbol represents the inverse of a statement and inverse of a conditional statement like   p → q is given as ~p → ~q which means if not p then not q
    Solution
    It is given
    p: Roses are red.
    q: Roses are beautiful.
    Symbolic statement of p → q is “ If Roses are red, then the roses are beautiful. ”
    Symbolic statement of ~q is “ Roses are not beautiful ”
    Symbolic statement of ~q → ~p is “ If Roses are not red, then the roses are not beautiful. ”
    Final Answer:
    p → q : If Roses are red, then the roses are beautiful
    ~q: Roses are not beautiful
    ~q → ~p: If Roses are not red, then the roses are not beautiful.
    General
    Maths-

    Make a valid conclusion in the situation.
    If cost price > selling price, then the transaction suffers loss.
    Cost price = $255 and selling price = $230.

    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Cost price > Selling price
    q: The transaction suffers loss
    So we can write the given statement “If cost price > selling price, then the transaction suffers loss” as:
    not stretchy rightwards arrowq
    We are given
    Cost price = $255 and Selling price = $230
    Here, Cost price > Selling price so we can say that the p statement is true and hence we can conclude that the q statement is also true i .e the transaction suffers loss.
    Final Answer:
    Hence we can conclude that the transaction suffers loss.

    Make a valid conclusion in the situation.
    If cost price > selling price, then the transaction suffers loss.
    Cost price = $255 and selling price = $230.

    Maths-General
    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Cost price > Selling price
    q: The transaction suffers loss
    So we can write the given statement “If cost price > selling price, then the transaction suffers loss” as:
    not stretchy rightwards arrowq
    We are given
    Cost price = $255 and Selling price = $230
    Here, Cost price > Selling price so we can say that the p statement is true and hence we can conclude that the q statement is also true i .e the transaction suffers loss.
    Final Answer:
    Hence we can conclude that the transaction suffers loss.
    parallel
    General
    Maths-

    Make a valid conclusion in the situation.
    If cost price < selling price, then the transaction makes profit.
    Cost price = $100 and selling price = $150.

    Hint:
    Law of Detachment states that if pnot stretchy rightwards arrow  q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Cost price < Selling price
    q: The transaction makes profit
    So we can write the given statement “If cost price < selling price, then the transaction makes profit” as:
    not stretchy rightwards arrowq
    We are given
    Cost price = $100 and Selling price = $150
    Here, Cost price < Selling price so we can say that the p statement is true and hence we can conclude that the q statement is also true i .e the transaction makes profit.
    Final Answer:
    Hence we can conclude that the transaction makes profit.

    Make a valid conclusion in the situation.
    If cost price < selling price, then the transaction makes profit.
    Cost price = $100 and selling price = $150.

    Maths-General
    Hint:
    Law of Detachment states that if pnot stretchy rightwards arrow  q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: Cost price < Selling price
    q: The transaction makes profit
    So we can write the given statement “If cost price < selling price, then the transaction makes profit” as:
    not stretchy rightwards arrowq
    We are given
    Cost price = $100 and Selling price = $150
    Here, Cost price < Selling price so we can say that the p statement is true and hence we can conclude that the q statement is also true i .e the transaction makes profit.
    Final Answer:
    Hence we can conclude that the transaction makes profit.
    General
    Maths-

    Make a valid conclusion in the situation.
    If you have more than $1000, then you can buy a music system.
    You have $1200.

    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow  q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: You have more than $1000
    q: You can buy a music system
    So we can write the given statement “If you have more than $1000, then you can buy a music system” as:
    not stretchy rightwards arrowq
    We are given that we have $1200 and $1200 > $1000 so we can say that the p statement is true and hence we can conclude that the q statement is also true i.e we can buy a music system.
    Final Answer:
    Hence we can conclude that we can buy a music system.

    Make a valid conclusion in the situation.
    If you have more than $1000, then you can buy a music system.
    You have $1200.

    Maths-General
    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow  q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: You have more than $1000
    q: You can buy a music system
    So we can write the given statement “If you have more than $1000, then you can buy a music system” as:
    not stretchy rightwards arrowq
    We are given that we have $1200 and $1200 > $1000 so we can say that the p statement is true and hence we can conclude that the q statement is also true i.e we can buy a music system.
    Final Answer:
    Hence we can conclude that we can buy a music system.
    General
    Maths-

    Make a valid conclusion in the situation.
    If 𝑛 < 5, then (−𝑛) > (−5).
    The value of 𝑛 is 3.

    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow  q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: 𝑛 < 5
    q: (−𝑛) > (−5)
    So we can write the given statement “If 𝑛 < 5, then (−𝑛) > (−5)” as:
    not stretchy rightwards arrow  q
    We are given
    n = 3
    Here, 3 < 5 so we can say that the p statement is true and hence we can conclude that the q statement is also true i.e (−3) > (−5).
    Final Answer:
    Hence we can conclude that (−3) > (−5).

    Make a valid conclusion in the situation.
    If 𝑛 < 5, then (−𝑛) > (−5).
    The value of 𝑛 is 3.

    Maths-General
    Hint:
    Law of Detachment states that if p not stretchy rightwards arrow  q is true and it is given that p is true then we can conclude that q is also true. Here, the statement is termed as Hypothesis and the statement q is termed as conclusion.
    Solution
    Consider the statement into two separate statements
    p: 𝑛 < 5
    q: (−𝑛) > (−5)
    So we can write the given statement “If 𝑛 < 5, then (−𝑛) > (−5)” as:
    not stretchy rightwards arrow  q
    We are given
    n = 3
    Here, 3 < 5 so we can say that the p statement is true and hence we can conclude that the q statement is also true i.e (−3) > (−5).
    Final Answer:
    Hence we can conclude that (−3) > (−5).
    parallel

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