Answer:
y = 3x + 12
Step-by-step explanation:
Slope (m) = 24-9/4-(-1)
= 25/5
= 3
Using point (-1 , 9)
y - y1 = m (x - x1)
y - 9 = 3 (x - -1)
y - 9 = 3 (x + 1)
y - 9 = 3x + 3
y = 3x + 3 + 9
y = 3x + 12
Answer:
Possible C:
(1 , 1 + 3sqrt(3))
(1 , 1 - 3sqrt(3))
Step-by-step explanation:
AB = 6 units
Midpoint of AB is (1,1)
C would be at x = 1
So, (1,y)
(y-1)² + (1+2)² = 6²
(y-1)² = 36 - 9
(y-1)² = 27
y - 1 = 3sqrt(3)
y - 1 = -3sqrt(3)
y = 1 + 3sqrt(3)
y = 1 - 3sqrt(3)
Answer:
The distance between A and B is 8.5 after rounded to the nearest tenth
Step-by-step explanation:
The formula you always use for distance between points is as pictured.
Point A is (-6, 4), this is x1 and y1
Point B is (2, 1), this is x2 and y2
(2-[-6])^2 + (1-4)^2
since 2 negatives make a positive, 2 - -6 = 8
1 - 4 = -3
This makes 8 squared and -3 squared
8x8 = 64
-3x-3 = 9
64 + 9 = 73
the square root of 73 = 8.544003745
After rounding to the nearest tenth, it is 8.5
I hope this helps you! Good luck :)
so, we have two 54x18 rectangles, so their perimeter is simply all those units added together, 54+54+54+54+18+18+18+18 = 288.
we know the circle's diameter is 1.5 times the width, well, the width is 18, so the diameter of the circle must be 1.5*18 = 27.
![\bf \stackrel{\textit{circumference of a circle}}{C=d\pi }~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ d=27 \end{cases}\implies C=27\pi \implies C=\stackrel{\pi =3.14}{84.78} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perimeter of the rectangles}}{288}~~~~+~~~~\stackrel{\textit{perimeter of the circle}}{84.78}~~~~=~~~~372.78](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bcircumference%20of%20a%20circle%7D%7D%7BC%3Dd%5Cpi%20%7D~~%20%5Cbegin%7Bcases%7D%20d%3Ddiameter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20d%3D27%20%5Cend%7Bcases%7D%5Cimplies%20C%3D27%5Cpi%20%5Cimplies%20C%3D%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B84.78%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20rectangles%7D%7D%7B288%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20circle%7D%7D%7B84.78%7D~~~~%3D~~~~372.78)
Answer:
0.84 square in
Step-by-step explanation:
Since the capacity of the cable is proportional to its cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold
2500 / 0.3 = 8333.33 lb/in
To old 7000 lb it the cross-sectional area would need to be
7000 / 8333.33 = 0.84 square in