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3 years ago

11

Find the area of 2/5 yd and 3/7 yd

1 answer:

mezya [45]3 years ago

4 0

(2/5)(3/7) = 6/35

The area of your square is 6/35 yds²

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(a+4)(a-2) please help

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Answer:

a²+2a-8

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

3 years ago

A triangle has sides measuring 7 centimeters and 13 centimeters that form an angle measuring 44°. Which of these is CLOSEST to t

Oliga [24]

Answer:

9.3 centimeters

Step-by-step explanation:

If the length of the third side of the triangle is c, then according to the Law of Cosines, c2 = 72 + 132 – 2(7)(13)(cos 44°). The value of c2 can be simplified to approximately 49 + 169 – 130.9198, or 87.0802, using a calculator. By taking the square root of 87.0802, the length of the third side of the triangle comes out to be approximately 9.3 centimeters.

8 0

3 years ago