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Verdich [7]
2 years ago
13

Enter an

Mathematics
1 answer:
bekas [8.4K]2 years ago
6 0

Answer:

9 - n

Step-by-step explanation:

could you make me brainliest if this helped? :)

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Find a quotient and remainder for 52/8
WITCHER [35]
52/8=6 remainder:4

8*6=48
52-48=4

Hope this really helps u out!!! : )
6 0
3 years ago
Read 2 more answers
Please help me out :/
tensa zangetsu [6.8K]
Add up all of your X’s!! then Try and add up all of the degrees I think. Check me if i’m wrong !
7 0
3 years ago
What is two and two fourths plus one third
kkurt [141]

Answer:

2 5/6

Step-by-step explanation:

2 2/4 + 1/3

First, you have to make the bottoms of the fraction the same, by figuring out the lowest common denominator. In this case it would be 12. 4x3 = 12. 3x4 = 12.

Multiply the top number by the same number you multiplied the bottom by.

We multiplied the 4 by 3, so we would also multiply the 2 by 3 (which would be 6).

Do the same for the second fraction. 1x4 = 4.

Now we have 2 6/12 + 4/12. We add the top numbers together and we get 10/12.

Now we have to reduce the fractions. We can do this in this situation by just dividing the top and bottom numbers by 2.

2 5/6

8 0
3 years ago
Please help !!!!!!!!!!!!!​
arlik [135]

Answer:

Step-by-step explanation:

7 0
3 years ago
16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.
Katyanochek1 [597]
We will use the sine and cosine of the sum of two angles, the sine and consine of \frac{\pi}{2}, and the relation of the tangent with the sine and cosine:

\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta

\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0

\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}

If you use those identities, for \alpha=x,\ \beta=\dfrac{\pi}{2}, you get:

\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x

\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x

Hence:

\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x
3 0
3 years ago
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