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3 years ago

Please try and hurry i need to finish fast

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

It is the one you selected

Step-by-step explanation:

To convert this into a fraction, you would have to move the "3" to the fraction. But the position of the "3" would ALWAYS be below the number which in this case would be "7" so it would be x ^ 7/3

plz brainliest :'D

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Geometry math question no Guessing and Please show work thank you

nasty-shy [4]

We can name an angle either by naming its vertex or by three letters , keeping the letter of vertex in between.

so here vertex is F

we can name it as ∠EFG, ∠F , ∠GFE

This strikes out the option B that is ∠G

so option B is the answer

6 0

3 years ago

Which expression is a difference of cubes? 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15- a^26

LiRa [457]

we know that

A polynomial in the form $a^{3}-b^{3}$ is called adifference of cubes. Both terms must be a perfect cubes

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

case A) $9w^{33} -y^{12}$

we know that

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

$w^{33}=(w^{11})^{3}$ ------> 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

case B) $18p^{15} -q^{21}$

we know that

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

$p^{15}=(p^{5})^{3}$ ------> 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

case C) $36a^{22} -b^{16}$

we know that

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

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

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

therefore

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

case D) $64c^{15} -a^{26}$

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

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

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

case E) $64c^{15} -d^{27}$

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

$64c^{15} -d^{27}=((2^{2})(c^{5}))^{3}-(d^{9})^{3}$

therefore

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

5 0

3 years ago

Read 2 more answers

Determine the Type I error if the null hypothesis, H0, is: an electrician claims that no more than 10% of homes in the city are

kap26 [50]

Answer : There are no more than 10% of homes in the city are not up to the current electric codes but actually we do not have evidence to accept it.

Explanation:

The Type I error occurs when the null hypothesis is true but we reject it.

Given null hypothesis : $H_0$ : an electrician claims that no more than 10% of homes in the city are not up to the current electric codes.

The Type I error for the given situation is that there are no more than 10% of homes in the city are not up to the current electric codes but actually we do not have evidence to accept it.

3 0

3 years ago

Hi there can anyone help me out with this please

DENIUS [597]

4.8*5.77 = 27.698km
27.698km to hectares is 2769.8 ha

8 0

4 years ago

Solve 3- x/2>_15 A. XS-24 B. X2-24 C. X2-6 D. XS 6

ch4aika [34]

Answer: x ≥ -24

Step-by-step explanation:

I think the equation is supposed to be written as an inequality.

3 - x/2 ≥ 15

Step 1: Subtraction property of equality

-x/2 ≥ 12

Step 2: Multiplication property of equality

-x ≥ 24

Step 3: Multiplication property of equality

x ≥ -24

8 0

3 years ago