The answer is the last one option: y+6=-34(x+2). I told you the reason in the other similar question.
Answer:
The answer to your question is:
x = 13
x + 10 = 23
5x - 42 = 23
Step-by-step explanation:
Data
angle 1 = x + 10
angle 2 = 5x - 42
They are vertical angles so, they measure the same.
x + 10 = 5x -42
x - 5x = -42 - 10
- 4x = -52
x = -52/-4
x = 13
angle 1 = x + 10 = 13 + 10 = 23
angle 2 = 5x - 42 = 5(13) -42 = 65 -42 = 23
Answer:
27/35 is the exact answer
7/10 is an estimate
Step-by-step explanation:
4/7+2/10
Simplify the fraction
4/7+1/5
The common denominator is 35
4/7*5/5 + 1/5 *7/7
20/35 + 7/35
27/35
To estimate
4/7 is close to 1/2
2/10 is close is 1/5
1/2 + 1/5 =
5/10 + 2/10 = 7/10
Answer:
A. Obtuse
Step-by-step explanation:
Well we can cross out C and D just by looking at the triangle because the triangle is not equal and there is no right angle.
Also it is not B an acute because not all the angles are acute.
Hence, the answer is A. Obtuse
the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.
from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.
the standing up sides are simply rectangles of 8x3.
if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D5%5C%5C%20p%3D24%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%285%29%2824%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bjust%20for%20one%20octagon%7D%7D%7BA%3D60%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwo%20octagon%27s%20area%7D%7D%7B2%2860%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Beight%20rectangle%27s%20area%7D%7D%7B8%283%5Ccdot%208%29%7D%5Cimplies%20120%2B192%5Cimplies%20312)
