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3 years ago

Find the product: Negative StartFraction 4 over 9 EndFraction and Negative StartFraction 3 over 8 EndFraction

Mathematics

2 answers:

kobusy [5.1K]3 years ago

8 0

Answer:

c 1/6

Step-by-step explanation:

been there done that

White raven [17]3 years ago

3 0

Answer:

1/6

Step-by-step explanation:

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(4v^3-3v^2)+(3v^3+6v^2-9v)

Alexxandr [17]

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7v^3+3v^2-9v

Step-by-step explanation:

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4 years ago

What is the distance between (9,5) and (-2,2)

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given that sin theta= 1/4, 0 is less than theta but less than pi/2, what is the exact value of cos theta

lapo4ka [179]

Answer:

$\cos{\theta} = \frac{\sqrt{15}}{4}$

Step-by-step explanation:

For any angle $\theta$ , we have that:

$(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1$

Quadrant:

$0 \leq \theta \leq \frac{\pi}{2}$ means that $\theta$ is in the first quadrant. This means that both the sine and the cosine have positive values.

Find the cosine:

$(\sin{\theta})^{2} + (\cos{\theta})^{2} = 1$

$(\frac{1}{4})^{2} + (\cos{\theta})^{2} = 1$

$\frac{1}{16} + (\cos{\theta})^{2} = 1$

$(\cos{\theta})^{2} = 1 - \frac{1}{16}$

$(\cos{\theta})^{2} = \frac{16-1}{16}$

$(\cos{\theta})^{2} = \frac{15}{16}$

$\cos{\theta} = \pm \sqrt{\frac{15}{16}}$

Since the angle is in the first quadrant, the cosine is positive.

$\cos{\theta} = \frac{\sqrt{15}}{4}$

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3 years ago

Please help meee, I’m having trouble in math. I solved the answer but got none of these.

tino4ka555 [31]

The answer is C

2.5 + 3.4 = 5.9

So:

5.9 x 10^4

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3 years ago