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gtnhenbr [62]
3 years ago
6

4 + 9 = A)49 B)14 C)none of the above(hint its A)

Mathematics
2 answers:
Mazyrski [523]3 years ago
8 0

Answer:

A

Step-by-step explanation:

elena-14-01-66 [18.8K]3 years ago
3 0

Answer:

a

Step-by-step explanation:

Well, I guess it's A, since you told me

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Verify sin4 x- sin2x = cos4x – cos2x is an identity
Andrej [43]

Step-by-step explanation:

sin⁴ x − sin² x

Factor:

(sin² x) (sin² x − 1)

Pythagorean identity:

(1 − cos² x) (-cos² x)

Distribute:

-cos² x + cos⁴ x

cos⁴ x − cos² x

6 0
3 years ago
Helppppppp pleaseeee
pochemuha

Answer:

-5

Step-by-step explanation:

a= -4 than 2 times-4. than be 8.and 4 square is -16. So it be 3+8-16 = -5

5 0
2 years ago
Read 2 more answers
Which of the following describes the square root of 41. 5,6 6,7 20,21 40,42
cestrela7 [59]

Answer:

6,7

Step-by-step explanation:

the squre root of 41 is 6.403

3 0
3 years ago
Someone please help me ASAP!
loris [4]
It would be (-5,-1) yw
8 0
3 years ago
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Find derivative problem<br> Find B’(6)
dalvyx [7]

Answer:

B^\prime(6) \approx -28.17

Step-by-step explanation:

We have:

\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]

We can move the constant outside:

\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]

Now, we will utilize the product rule. The product rule is:

(uv)^\prime=u^\prime v+u v^\prime

We will let:

\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t

Then:

\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1

(The derivative of u was determined using the chain rule.)

Then it follows that:

\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}

Therefore:

\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]

By simplification:

\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0
3 years ago
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