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AveGali [126]
3 years ago
6

Find sin (a) given that cos (a)=1

Mathematics
1 answer:
S_A_V [24]3 years ago
4 0

Solve the trigonometric equation by isolating the function and then taking the inverse. Use the period to find the full set of all solutions.

a=2πn, for any integer n

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...........I need help......
ASHA 777 [7]

The anwer is B, f(x)=9500-750x

Hope this helps, brianliest if you can!

5 0
4 years ago
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
Point w id located at (-2,3) on a coordinate plane. Point W is reflected over the x-axis to create point W" is then reflected ov
Anna11 [10]

Answer:

W'' = (2, -3)

Step-by-step explanation:

Reflection over the x-axis negates the y-coordinate and leaves the x-coordinate alone. The W becomes ...

... W' = (-2, -3)

Reflection over the y-axis negates the x-coordinate and leaves the y-coordinate alone. The W' becomes ...

... W'' = (2, -3)

5 0
3 years ago
Read 2 more answers
I need help with this .
vaieri [72.5K]
I would be glad to help you! I will solve through 3 to get you started.

This is a very important concept, so it would be helpful if you learn it on your own.

C= 2* pi* r

1. The radius is 5 cm. 5 cm represents r in the equation.
C= 2* 3.14 (or pi)* 5 
Multiply it by using a calculator. 
My answer was 31.4.

Let's try the second one.
2. The diameter is 9. Have of the diameter is the radius. 
9/2= 4.5 
Plug it into the equation.
C= 2*3.14*4.5
I got 28.26

One more. 
Plug it into the equation.
C= 2*3.14*5.6
35.17 was my answer.

Make sure to add the end signs to you answer, ft, cm, meter, etc.

Let me know if you need more help.

Brainliest answer is always appreciated.
7 0
3 years ago
Compute the dimensions of gravitational constant G where F=Gm1m2/r^2​
VARVARA [1.3K]

Answer:

G=L^3/(T^2M)

Step-by-step explanation:

  1. G=(Fr^2)/Mm
  2. G=(ML^3)/(T^2M^2)
  3. G=L^3/(T^2M)
8 0
3 years ago
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