Add 4.5<span> and </span>9.2<span> to get </span><span>13.7.
</span>13.7
13.7 in. is the length
Answer:
C

Co efficient means the number or value beside preventing the variable from standing alone
To find the x-intercept, you simply make y equal to 0 and solve the equation for x. To find the y-intercept, you make x equal to 0 and solve for y. So, for number 1, the answer is D because when you set it to y equals 0, you get x=3 and when you set it to x equals 0, you get 4.
Number 2 we would do the same thing and get D as well.
Hope this helps!
Answer:
1 in 47.
Step-by-step explanation:
Add all of the outcomes (28 + 2 + 3 + 14).
You'll get 47.
Do not add the 1s.
There is your answer.
1 in 47.
Answer: 
<u>Step-by-step explanation:</u>
(4, 1) & (2, -5)
First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

Next, find the midpoint of (4, 1) and (2, -5):

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:
