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bogdanovich [222]

3 years ago

6

Which is not a like term in the expression 7x +3 - 4 x-x

1 answer:

777dan777 [17]3 years ago

7 0

Answer:

3 is not a like term because it isn't an x term, it doesn't have an x variable while the other terms do.

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1000 grams decreased by 94%

gregori [183]

$subtract\ 94\%\ of\ 1000\ from\ 1000\\\\1000-94\%*10000=1000-0.94*1000=1000-940=60\\\\
1000\ grams\ decreased\ by\ 94\%\ is\ 60.$

3 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

(6x4 – 6х3 + 11х2 -х + 20) /<br> (3х2 + 3х + 4)

Fittoniya [83]

Answer:

6638_729_88 and so this is what thay saif was right i leterkly just had to answer this

6 0

2 years ago

Multiply and simplify: 3a(4a2 – 5a + 12)

DochEvi [55]

3a(4a²-5a+12)

12a³-15a²+36a

~Hope this helped!~

3 0

3 years ago

Find the variance of: 6 6 7 8 10 10

marta [7]

<h3>3.368</h3>

Step-by-step explanation:

I hope this helps!!!!!

7 0

2 years ago