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Pachacha [2.7K]
3 years ago
7

Which equation is NOT a linear function?

Mathematics
1 answer:
Irina18 [472]3 years ago
4 0

Answer:

the 3rd one is not linear

y= x to the 2nd power

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What is the mean of the following set of numbers?
amm1812

Answer:

C. 279.3

Step-by-step explanation:

To find the mean of a set of numbers you have to add all of them together and divide them by how manys numbers there are

If we add all the numbers together it equals 2514

There are 9 numbers here

2514/9=279,333333

Therefore, the answer is C

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3 years ago
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Flynn has $4 worth of quarters.
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16 quarters

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Flynn has 16 quarters.

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PLEASE HELP ASAP
LenaWriter [7]

Answer:

Optiom B

Step-by-step explanation:

If you use sine rule here, you'll find the other two sides

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3 years ago
Private company wanted to donate black shoes for the students for the upcoming school year
VMariaS [17]

Two possible questions are:

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<h3>What are the two questions?</h3>

The appropriate questions are questions that can be solved using  either statistics or mathematical operation.

For example, In order to determine the total cost of the donation, the total shoes donated would have to multiplied by the cost per shoe.

Here is the complete question?
Directions: Read and analyze the situation below and write 2 possible

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1. A private company wanted to donate black shoes for the students for

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To learn more about statistics, please check: brainly.com/question/19628082

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2 years ago
Prove-: sin5A = 5cos^4 A sinA - 10cos^2 A sin^3 A + sin^5 A
jeyben [28]
A=x\\\\sin5x=5cos^4xsinx-10cos^2xsin^3x+sin^5x\\\\L=sin(3x+2x)=sin3xcos2x+sin2xcos3x=(*)\\-------------------------------\\sin3x=sin(2x+x)=sin2xcosx+sinxcos2x\\=2sinxcosxcosx+sinx(cos^2x-sin^2x)\\=2sinxcos^2x+sinxcos^2x-sin^3x=3sinxcos^2x-sin^3x\\-------------------------------

sin3xcos2x=(3sinxcos^2x-sin^3x)(cos^2x-sin^2x)\\=3sinxcos^4x-3sin^3xcos^2x-sin^3xcos^2x+sin^5x\\=3sinxcos^4x-4sin^3xcos^2x+sin^5x\\-------------------------------

cos3x=cos(2x+x)=cos2xcosx-sin2xsinx\\=cosx(cos^2x-sin^2x)-2sinxcosxsinx\\=cos^3x-sin^2xcosx-2sin^2xcosx=cos^3x-3sin^2xcosx\\-------------------------------

sin2xcos3x=2sinxcosx(cos^3x-3sin^2xcosx)\\=2sinxcos^4x-6sin^3xcos^2x\\-------------------------------

&#10;(*)=3sinxcos^4x-4sin^3xcos^2x+sin^5x+2sinxcos^4x-6sin^3xcos^2x\\\\=5sinxcos^4x-10sin^3xcos^2x+sin^5x=R

=======================================\\\\sin(\alpha+\beta)=sin\alpha cos\beta+sin\beta cos\alpha\\\\cos(\alpha+\beta)=cos\alpha cos\beta-sin\alpha sin\beta\\\\sin2\alpha=2sin\alpha cos\alpha\\\\cos\alpha=cos^2\alpha-sin^2\alpha
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3 years ago
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