25% of 88 =
0.25 × 88 = 22
88 + 22 = 110
Answer = 110
Hope this helped☺☺
Answer:
Yes
Step-by-step explanation:
The formula for area of a triangle is A = (1/2)bh,
For the first triangle we can leave it in general terms, so it's area is
A = (1/2)bh, depending on what b and h are, but it doesn't matter here...
The second triangle has base that is twice the other triangles base. Bases being multiples of each other is the definition of being proportional so the bases are proportional, an the area of the second triangle is
A = (1/2)(2b)h, which simplifies to
A = bh
Comparing the 2 areas, you can see that one has a multiplier of (1/2), so their areas are proportional
<span>3x+2y+4+9y=
Combine like terms:
2y+9y=11y
I believe the answer is 3x+11y+4
(That is unless you know the numbers that y & x equals)</span>
if the diameter is 20, the its radius must be half that or 10.
![\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta \pi r^2}{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{degrees}{angle}\\[-0.5em] \hrulefill\\ A=5\pi \\ r=10 \end{cases}\implies \begin{array}{llll} 5\pi =\cfrac{\theta \pi (10)^2}{360}\implies 5\pi =\cfrac{5\pi \theta }{18} \\\\\\ \cfrac{5\pi }{5\pi }=\cfrac{\theta }{18}\implies 1=\cfrac{\theta }{18}\implies 18=\theta \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20sector%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%5E2%7D%7B360%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3D%5Cstackrel%7Bdegrees%7D%7Bangle%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20A%3D5%5Cpi%20%5C%5C%20r%3D10%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%205%5Cpi%20%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20%2810%29%5E2%7D%7B360%7D%5Cimplies%205%5Cpi%20%3D%5Ccfrac%7B5%5Cpi%20%5Ctheta%20%7D%7B18%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B5%5Cpi%20%7D%7B5%5Cpi%20%7D%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%201%3D%5Ccfrac%7B%5Ctheta%20%7D%7B18%7D%5Cimplies%2018%3D%5Ctheta%20%5Cend%7Barray%7D)
Answer:
D
Step-by-step explanation:
First add all the simple line up so;
Then use Pythagoras to work out the lengths of the remaining lines
So to three decimal places it would be;
