Ask question

Ask question

All categories

3 years ago

10

What is the multiplicity of each of the roots of the graph of

2 answers:

skelet666 [1.2K]3 years ago

7 0

Answer:

C

Step-by-step explanation:

pantera1 [17]3 years ago

3 0

Answer:

The correct option is;

C. -3, multiplicity 2; -1, multiplicity 1; 1, multiplicity 1

Please find attached the required function graph

Step-by-step explanation:

To solve the equation f(x) = 2·x⁴ + 12·x³ + 16·x² -12·x - 18, by graphing the function, we have;

x ${}$ F(x)

-4 ${}$ 30

-3 ${}$ 0

-2 ${}$ 6

-1 ${}$ 0

0 ${}$ -18

1 ${}$ 0

2 ${}$ 150

The shape of a graph with multiplicity of 2

Given that the graph bounces of the horizontal axis at the y-intercept at point x = -3, the factor (x - 3) must be a quadratic of the form (x - 3)², thereby having a multiplicity of 2 in the solution which are;

x = 1, -1, and, giving

(x - 1)·(x + 1)·(x - 3)² = 0

Therefore, the correct option is -3, multiplicity 2; -1, multiplicity 1; 1 multiplicity 1.

You might be interested in

Algebra 1 ) U. Checkpoint: Systems of equations and inequalities LOW

krek1111 [17]

Answer:algebra ok

Step-by-step explanation:

7 0

3 years ago

Which inequality best represents the graph? O A. 2x + y = 6 O B. x + 2y 6 O c. 2r- y< 6 O D. 2x - y > -6 E. 2x + y26

Misha Larkins [42]

Answer:

It would be A

Step-by-step explanation:

5 0

3 years ago

When you look at the two cylinders with these values of rand h, what do you notice about the proportions of the cylinders?

Klio2033 [76]

Answer:

One of the cylinders is short and wide, while the other is tall and thin.

Step-by-step explanation:

7 0

3 years ago

Simplify 3√ ̅10 + 7 √ ̅15 - 6 √ ̅10 - 4 √ ̅15

skelet666 [1.2K]

Answer:

-3√ ̅10+3 √ ̅15

Step-by-step explanation:

7 0

3 years ago

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

4 0

4 years ago