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Alina [70]
2 years ago
7

Pls Solve my question ​

Mathematics
1 answer:
JulsSmile [24]2 years ago
7 0

hope this will help you more

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Dylan walked 1/2 miles in 20 minutes. How far will he go (in miles) in 120 minutes? ​
viva [34]

Answer:

He will have gone 3 miles after 120 minutes

Step-by-step explanation: 120/20=6        1/2 times 6 is 3.

4 0
3 years ago
Read 2 more answers
An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu
exis [7]
Take the derivative with respect to t
- w \sin(wt) + \sqrt{8} w cos(wt)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
0 = -w \sin(wt) + \sqrt{8} w cos(wt)
divide by w
0 =- \sin(wt) + \sqrt{8} cos(wt)
we add sin(wt) to both sides

\sin(wt)= \sqrt{8} cos(wt)
divide both sides by cos(wt)
\frac{sin(wt)}{cos(wt)}= \sqrt{8}   \\  \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\  \\ wt=arctan(2 \sqrt{2)} OR\\ wt=arctan( { \frac{1}{ \sqrt{2} } )
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

I=cos(2n \pi -2arctan( \sqrt{2} ))
since 2npi is just the period of cos
cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 

substituting our second soultion we get
I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))
since 2npi is the period
I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}
so the maximum value =\frac{1}{3}
minimum value =- \frac{1}{3}


4 0
3 years ago
Anwser both of them please
eimsori [14]
The rectangle on the right is 81 and the rectangle on the left is 24. I don't really know about the second one sorry, but I hope this helped.
4 0
3 years ago
Jayla bought a table and some chairs at a furniture store.
anzhelika [568]

Answer:

b. She paid $30 for each chair.

6 0
3 years ago
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In circle O, central angle AOB has a measure of 90°. Which of the following is not true? segment OA is a radius segments OA and
lesantik [10]
The last one: the arc AB is a semicircle. 

a semicircle is 180 degrees
3 0
3 years ago
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