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3 years ago

6

4(5+3)-(8/4)+34 what is answer

1 answer:

levacccp [35]3 years ago

3 0

<span>4*(5+3)-(8/4)+3*4

4(8) - (2) + 3 x 4

32 - 2 + 12

30 + 12

42

The answer is 42, I believe.</span>

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2 years ago

Read 2 more answers

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

Paha777 [63]

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,

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

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

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

We know that,

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

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

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

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

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

Using the trigonometric ratios, we get

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

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

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

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

6 0

3 years ago

Which value for x makes this equation true? 2x + 5 = 9 *<br> -4<br> 4<br> 2<br> -2

masya89 [10]

2x + 5 = 9
2x +(5-5)= 9-5
2x= 4
2x/2= 4/2

X= 2

3 0

3 years ago