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pickupchik [31]

3 years ago

9

What are the solutions of this equation 16-2x^2=-64

1 answer:

xenn [34]3 years ago

5 0

Answer:

x = ± 2 $\sqrt{10}$

Step-by-step explanation:

Given

16 - 2x² = - 64 ( subtract 16 from both sides )

- 2x² = - 80 ( divide both sides by - 2 )

x² = 40 ( take the square root of both sides )

x = ± $\sqrt{40}$ = ± $\sqrt{4(10)}$ = ± 2 $\sqrt{10}$

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2 years ago

Given sin(u)= -7/25 and cos(v) = -4/5, what is the exact value of cos(u-v) if both angles are in quadrant 3

solmaris [256]

Given:

$\sin (u)=-\dfrac{7}{25}$

$\cos (v)=-\dfrac{4}{5}$

To find:

The exact value of cos(u-v) if both angles are in quadrant 3.

Solution:

In 3rd quadrant, cos and sin both trigonometric ratios are negative.

We have,

$\sin (u)=-\dfrac{7}{25}$

$\cos (v)=-\dfrac{4}{5}$

Now,

$\cos (u)=-\sqrt{1-\sin^2 (u)}$

$\cos (u)=-\sqrt{1-(-\dfrac{7}{25})^2}$

$\cos (u)=-\sqrt{1-\dfrac{49}{625}}$

$\cos (u)=-\sqrt{\dfrac{625-49}{625}}$

On further simplification, we get

$\cos (u)=-\sqrt{\dfrac{576}{625}}$

$\cos (u)=-\dfrac{24}{25}$

Similarly,

$\sin (v)=-\sqrt{1-\cos^2 (v)}$

$\sin (v)=-\sqrt{1-(-\dfrac{4}{5})^2}$

$\sin (v)=-\sqrt{1-\dfrac{16}{25}}$

$\sin (v)=-\sqrt{\dfrac{25-16}{25}}$

$\sin (v)=-\sqrt{\dfrac{9}{25}}$

$\sin (v)=-\dfrac{3}{5}$

Now,

$\cos (u-v)=\cos u\cos v+\sin u\sin v$

$\cos (u-v)=\left(-\dfrac{24}{25}\right)\left(-\dfrac{4}{5}\right)+\left(-\dfrac{7}{25}\right)\left(-\dfrac{3}{25}\right)$

$\cos (u-v)=\dfrac{96}{625}+\dfrac{21}{625}$

$\cos (u-v)=\dfrac{1 17}{625}$

Therefore, the value of cos (u-v) is 0.1872.

6 0

2 years ago