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MatroZZZ [7]
3 years ago
14

Slove for inequality of -6> t-(-13)

Mathematics
2 answers:
Tpy6a [65]3 years ago
4 0

Step-by-step explanation:

-6>t-(-13)

= -6>t+13

= -6-13>t

= -19>t

= t<-19

zlopas [31]3 years ago
3 0

Answer:

t < - 19

Step-by-step explanation:

Given

- 6 > t - (- 13) , that is

- 6 > t + 13 ( subtract 13 from both sides )

- 19 > t , then

t < - 19

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Easy Math, please help!
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Answer:

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Step-by-step explanation:

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8 0
3 years ago
 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
2 years ago
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tankabanditka [31]

cant help when you put question marks!

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8 0
3 years ago
Arsham is training for a race .after 15 minutes he has 22 laps left to run after 44 minutes he has 2 laps left what is the total
olga55 [171]
 Arsham ran 20 laps because 22- 2= 20.

3 0
3 years ago
Find the slope of the line which goes through the given points.<br> d<br> A(2,8) and B(2,6)
KATRIN_1 [288]

Answer:

Undefined

Step-by-step explanation:

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Hope this helps!

7 0
3 years ago
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