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ad-work [718]
3 years ago
11

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

Mathematics
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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Express 105/20 as a mixed number in simpelist form
Sedbober [7]

The correct answer to this question is

D  - 5  1/4

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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
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