Hello!
Since D is the midpoint and the two equations are and both sides of point D the equations equal each other
9x - 7 = 3x + 17
Now you solve it algebraically
Add 7 to both sides
9x = 3x + 24
Subtract 3x from both sides
6x = 24
Divide both sides by 6
x = 4
Now we put this into both equations and add them
9(4) - 7 = 29
3(4) + 17 = 29
29 + 29 = 58
The answer is 58 units
Hope this helps!
we have the function

Part a
For t=7
substitute in the given function

For t=14

For t=21

For t=28

For t=35

Observation: The values of E varies from -1 to 1, including the zero
Part B
Remember that
The Period goes from one peak to the next
so
Period=2pi/B
B=pi/14
Period=(2pi)/(pi/14)=2pi*14/pi=28
<h2>the period is 28 days</h2>
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)
Answer:
Expression: 0.15 * 20
Equation: 0.15 * 20 = 3
Step-by-step explanation:
This expression was found by dividing 15% by 100 to convert it into a decimal (0.15) that can be used to multiply with 20 to get your answer.