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sertanlavr [38]
3 years ago
10

Simplifyn-3 n-2/n^-6n^5A) 1/n^-4B)1/n^4C)n^-6D)n^4

Mathematics
1 answer:
pychu [463]3 years ago
5 0

Answer:

Step-by-step explanation: was uuu

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Acute angle DOG with a measure of 45 degrees
Vlad1618 [11]
So make and angle with the measure of 45 degrees. And then label it DOG. It doesn't matter where you start labeling as long as O is your midpoint. :)
6 0
3 years ago
F(x)=2x ​2 ​​ +5√ ​(x+2) ​ ​​f(2)=
irga5000 [103]

f(x) 2x² + 5√(x+2)


f(2) = 2(2)² + 5√(2+2)

     = 2(4) + 5√4

     = 8 + 10

     = 18      <======= answer

4 0
3 years ago
(4x3 + 3x2 - 30x - 10) ÷ (x -3)
pogonyaev
(12x+6x-30x-10)/(x-3)=
=(-12x-10)/(x-3)
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5 0
3 years ago
Verify the identity 6cos^2(x) -3 = 3 - 6sin^2(x)
Lorico [155]

Answer:

It is proved that  6\cos ^{2}x -3 = 3 - 6\sin ^{2} x .

Step-by-step explanation:

We already have the identity of x as \sin ^{2}x + \cos ^{2}x = 1 .......... (1)  .

So, from equation (1) we can write that

\cos ^{2} x = 1 - \sin ^{2} x

⇒ 6\cos ^{2} x = 6 - 6 \sin ^{2} x

⇒ 6\cos ^{2} x -3 = 6 - 6 \sin ^{2}x -3

⇒ 6\cos ^{2}x -3 = 3 - 6\sin ^{2} x

Hence, it is proved that  6\cos ^{2}x -3 = 3 - 6\sin ^{2} x . (Answer)

5 0
3 years ago
Please help me IDK how to do this!!!!!!
tatiyna
A parallelogram should have 2 sets of parallel lines. Let's find the slope of line PQ and RS to test.

PQ:
(4-2)/(1-(-3))
2/4
1/2

RS:
(2-0)/(3-1)
2/2
1

Because 1 does not equal 1/2 (the slopes are different) the lines are not parallel. Thus, the figure is not a parallelogram.
5 0
3 years ago
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