Answer:
0. toab1. The right answer for K12 is ( > ) I Took the test and is correct. 4×16-16 > 4× [24-2× (4+8)] heart outlined. Thanks 0. star outlined. star outlined. star outlined.
For a , C= 18.85
for b , C= 21.99
for 1 , C= 75.4
for 2 , C= 87.96
<h3>
Answer: Approximately 13 square units (choice B)</h3>
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Explanation:
The given reflex angle is 215 degrees. A reflex angle is anything over 180 degrees, but less than 360. Subtract 215 from 360 to get the measure of angle AOB
angle AOB = 360 - 215 = 145
angle AOB = 145 degrees
We'll use this later.
Now find the area of the full circle. Use the formula A = pi*r^2. The radius is r = sqrt(10) which can be found through the distance formula or the pythagorean theorem. You want to find the length of either OA or OB to get the radius.
The area of the circle is
A = pi*r^2
A = pi*(sqrt(10))^2
A = 10pi
This is the exact area of the full circle, but we want just a fractional portion of it. Specifically we want the pie slice that is formed by angle AOB
area of sector AOB = [ (angle AOB)/360 ] * (area of full circle)
area of sector AOB = (145/360)*10pi
area of sector AOB = 145pi/36
area of sector AOB = 145*3.14/36
area of sector AOB = 12.647 approximately
area of sector AOB = 13 square units approximately, after rounding to the nearest whole number
Answer:
8
Step-by-step explanation:
hope this is right????
im not sure if i did it correctly but i tryed
Check the picture below.
so by graphing those two, we get that little section in gray as you see there, now, x = 6 is a vertical line, so we'll have to put the equations in y-terms and this is a washer, so we'll use the washer method.
the way I get the radii is by using the "area under the curve" way, namely, I use it to get R² once and again to get r² and using each time the axis of rotation as one of my functions, in this case the axis of rotation will be f(x), and to get R² will use the "farthest from the axis of rotation" radius, and for r² the "closest to the axis of rotation".
now, both lines if do an equation on where they meet or where one equals the other, we'd get the values for y = 0 and y = 1, not surprisingly in the picture.
![\displaystyle\pi \int_0^1\left( 3y-3y^2-\cfrac{y^2}{16}+\cfrac{y^4}{16} \right)dy\implies \pi \left( \left. \cfrac{3y^2}{2} \right]_0^1-\left. y^3\cfrac{}{} \right]_0^1-\left. \cfrac{y^3}{48}\right]_0^1+\left. \cfrac{y^5}{80} \right]_0^1 \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{59\pi }{120}~\hfill](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cpi%20%5Cint_0%5E1%5Cleft%28%203y-3y%5E2-%5Ccfrac%7By%5E2%7D%7B16%7D%2B%5Ccfrac%7By%5E4%7D%7B16%7D%20%5Cright%29dy%5Cimplies%20%5Cpi%20%5Cleft%28%20%5Cleft.%20%5Ccfrac%7B3y%5E2%7D%7B2%7D%20%5Cright%5D_0%5E1-%5Cleft.%20y%5E3%5Ccfrac%7B%7D%7B%7D%20%5Cright%5D_0%5E1-%5Cleft.%20%5Ccfrac%7By%5E3%7D%7B48%7D%5Cright%5D_0%5E1%2B%5Cleft.%20%5Ccfrac%7By%5E5%7D%7B80%7D%20%5Cright%5D_0%5E1%20%5Cright%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Ccfrac%7B59%5Cpi%20%7D%7B120%7D~%5Chfill)
