For the first one: A=pi*r^2*1/8 (1/8 becuase we are finding the area of one pane)
A=pi*16/8
A=2pi
For the second one: First do the larger circle minus the smaller circle (pretend they are whole circles)
A(larger)=pi*18^2
=324pi
A(smaller)=pi*8^2
=64pi
A=324pi-64pi =260pi
Now, a whole circle is 360 degrees so 150 degrees is 5/12 of a circle. We take 5/12 of 260pi to get the answer
260pi*5/12=340.3
First we find the measurement of dark dot .
ANd for that we have to use the formula of the angle formed by the intersection of chord .
Let the dark dot be t.
So we have
![t =\frac{1}{2} (96+66)=81](https://tex.z-dn.net/?f=%20t%20%3D%5Cfrac%7B1%7D%7B2%7D%20%2896%2B66%29%3D81%20%20)
t and z are linear pair and sum of angles of linear pair =180 degree.
So we get
![t+z=180 \\ 81+z =180 \\ z = 180-81 =99 degree](https://tex.z-dn.net/?f=%20t%2Bz%3D180%20%5C%5C%2081%2Bz%20%3D180%20%5C%5C%20z%20%3D%20180-81%20%3D99%20degree%20)
The first picture about the rational numbers I am not sure the answer.
In the second picture the set that are all equivalent to the fraction is the second set.
In the third picture
the equivalent expression is the first one.
In the fourth picture the correct situation is the first one about buying popcorn.
Hope this helps a little, sorry I could not help more.
well, we know the ceiling is 6+2/3 high, and Eduardo has 4+1/2 yards only, how much more does he need, well, is simply their difference, let's firstly convert the mixed fractions to improper fractions and then subtract.
![\stackrel{mixed}{6\frac{2}{3}}\implies \cfrac{6\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{20}{3}} ~\hfill \stackrel{mixed}{4\frac{1}{2}}\implies \cfrac{4\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{9}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{20}{3}-\cfrac{9}{2}\implies \stackrel{using ~~\stackrel{LCD}{6}}{\cfrac{(2\cdot 20)-(3\cdot 9)}{6}}\implies \cfrac{40-27}{6}\implies \cfrac{13}{6}\implies\blacktriangleright 2\frac{1}{6} \blacktriangleleft](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%203%2B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B20%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B9%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B20%7D%7B3%7D-%5Ccfrac%7B9%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Busing%20~~%5Cstackrel%7BLCD%7D%7B6%7D%7D%7B%5Ccfrac%7B%282%5Ccdot%2020%29-%283%5Ccdot%209%29%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B40-27%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B13%7D%7B6%7D%5Cimplies%5Cblacktriangleright%202%5Cfrac%7B1%7D%7B6%7D%20%5Cblacktriangleleft)
Answer:
w = ![\frac{p -2l}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bp%20-2l%7D%7B2%7D)
Step-by-step explanation:
p = 2l + 2w Subtract 2l from both sides of the equation
p - 2l = 2p Divide both sides by 2
= w