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4 years ago

8

a survey of students in a school showed 1/2 of the students exercise regularly, 2/5 of the students exercise occasionally and th

e rest of the students do not exercise at all what fraction of the students do not exercise

1 answer:

Sveta_85 [38]4 years ago

5 0

Answer:

1/10

Step-by-step explanation:

add 2/5 and 1/2 then subtract from 1 whole to find the remainder

make it so they have the same denominator

2/5 = 4/10

1/2 = 5/10

4/10 + 5/10 = 9/10

1=10/10

10/10-9/10 = 1/10

so 1/10 of the students don't exercise at all

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we know that

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Let's verify each case to determine the solution to the problem

<u>case A)</u> $9w^{33} -y^{12}$

we know that

$9=3^{2}$ ------> <u>the term is not a perfect cube</u>

$w^{33}=(w^{11})^{3}$ ------> the term is a perfect cube

$y^{12}=(y^{4})^{3}$ ------> the term is a perfect cube

therefore

The expression $9w^{33} -y^{12}$ is not a difference of cubes because the term $9$ is not a perfect cube

<u>case B)</u> $18p^{15} -q^{21}$

we know that

$18=2*3^{2}$ ------> <u>the term is not a perfect cube</u>

$p^{15}=(p^{5})^{3}$ ------> the term is a perfect cube

$q^{21}=(q^{7})^{3}$ ------> the term is a perfect cube

therefore

The expression $18p^{15} -q^{21}$ is not a difference of cubes because the term $18$ is not a perfect cube

<u>case C)</u> $36a^{22} -b^{16}$

we know that

$36=2^{2}*3^{2}$ ------> <u>the term is not a perfect cube</u>

$a^{22}$ ------> <u>the term is not a perfect cube</u>

$b^{16}$ ------> <u>the term is not a perfect cube</u>

therefore

The expression $36a^{22} -b^{16}$ is not a difference of cubes because all terms are not perfect cubes

<u>case D)</u> $64c^{15} -a^{26}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$a^{26}$ ------> <u>the term is not a perfect cube</u>

therefore

The expression $64c^{15} -a^{26}$ is not a difference of cubes because the term $a^{26}$ is not a perfect cube

I'm adding a new case so I can better explain the problem

<u>case E)</u> $64c^{15} -d^{27}$

we know that

$64=2^{6}=(2^{2})^{3}$ ------> the term is a perfect cube

$c^{15}=(c^{5})^{3}$ ------> the term is a perfect cube

$d^{27}=(d^{9})^{3}$ ------> the term is a perfect cube

Substitute

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therefore

The expression $64c^{15} -d^{27}$ is a difference of cubes because all terms are perfect cubes

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