Y = mx + b
slope(m) = 7
(5,30)...x = 5 and y = 30
now we sub and find b, the y int
30 = 7(5) + b
30 = 35 + b
30 - 35 = b
-5 = b
so ur equation is : y = 7x - 5 <==
![\bf \stackrel{\textit{volume of a cylinder}}{V=\pi r^2 h}~\hspace{7em}\stackrel{\textit{volume of a cone}}{V=\cfrac{\pi r^2 h}{3}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{volume of a cylinder}}{V=24\pi }~\hspace{7em}\stackrel{\textit{volume of a cone}}{V=\cfrac{24\pi }{3}}\implies V=8\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%7D%7BV%3D%5Cpi%20r%5E2%20h%7D~%5Chspace%7B7em%7D%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20a%20cone%7D%7D%7BV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%7D%7BV%3D24%5Cpi%20%7D~%5Chspace%7B7em%7D%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20a%20cone%7D%7D%7BV%3D%5Ccfrac%7B24%5Cpi%20%7D%7B3%7D%7D%5Cimplies%20V%3D8%5Cpi)
notice the volumes, the cone's volume is really one-third that of the cylinder, assuming "h"eight and "r"adius is the same on both.
To find the area of the shaded piece, subtract the square area to the area of the sector.
<span>6 × 6 − <span>1/2</span><span>r^2 0</span></span>Subtract <span>☇
</span>36−1/2(6)2pi/2
☇
<span>36−<span>1/2</span>(6<span>)2</span><span><span>pi/</span>2
</span></span>☇
7.722 <span>✔
The area is 7.722. </span><span>✩</span>
Answer:
Amount to be needed = 3/8
Step-by-step explanation:
3/4 pound of dried apples to 1/2 pound of pears
3/4 : 1/2
Amount of dried apples to be needed per pound of dried pears = ??
Therefore, 3/4×1/2 = 3/8
Amount to be needed = 3/8
