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True [87]
3 years ago
15

Math question see if you're able to help!!!*plz explain*

Mathematics
1 answer:
kotykmax [81]3 years ago
5 0

I'm going to do this the long way, finding everything, but there are shorter methods if all you are trying to do is order n^2. I'm also going to start by removing the brackets in each question.

A

- 5n^3 + 5n^2 + 5 + n^3 - n^2

-4n^3 + 4n^2 + 5

B

n^3 + 2n^2 - n - 2 - (n^3 - 3n^2)

n^3 + 3n^2 - n - 3 - n^3 + 3n^2

5n^2 - n - 2

C

n^2(n - 4)+ 5n^3 - 6

n^3 - 4n^2 + 5n^3 - 6

6n^3 - 4n^2 - 6

D

2n^3 - 4n^2 - 2n + 3n^2

2n^3 - n^2 - 2n

The correct order is bad C or B A D C.

I don't know what to put into the boxes. You might have to ask about that.

The numbers in front of the coefficients for x^2 are

B = 5

A = 4

D = - 1

C = - 4

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<u><em>The complete question is</em></u>

Consider this right triangle. 21 29 20 Write the ratio equivalent to: Sin B - CscA- Cot B

The picture of the question in the attached figure

Part 1) Write the ratio equivalent to: Sin B

we know that

In the right triangle ABC

sin(B)=\frac{AC}{AB} ----> by SOH (opposite side divided by the hypotenuse)

substitute the values

sin(B)=\frac{21}{29}

Part 2) Write the ratio equivalent to: Csc A

we know that

In the right triangle ABC

csc(A)=\frac{1}{sin(A)}

sin(A)=\frac{BC}{AB} -----> by SOH (opposite side divided by the hypotenuse)

substitute the values

sin(A)=\frac{20}{29}

therefore

csc(A)=\frac{29}{20}

Part 3) Write the ratio equivalent to: Cot A

we know that

In the right triangle ABC

cot(A)=\frac{1}{tan(A)}

tan(A)=\frac{BC}{AC} -----> by TOA (opposite side divided by the adjacent side)

substitute the values

tan(A)=\frac{20}{21}

therefore

cot(A)=\frac{21}{20}

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