To solve this, you’ll first need to solve for their slopes.
The slope for line Q is y2-y1/x2-x1 = -8-(-2)/-8-(-10) = -3
We know that the lines are perpendicular so the negative reciprocal of -3 is 1/3
The equation you get it y = 1/3x + b.
Now you will need to solve for b by substituting in the first ordered pair of line R.
2 = 1/3(1) + b.
Once you solve for b, you should get 5/3 and y = 1/3x + 5/3
Now, to find a, you will need to substitute in 10 from the second ordered pair into x in your new equation.
y = 1/3(10) + 5/3.
Your solution should be 5.
So your answer is: a = 5
First you need to find the surface area of the wall that was painted, and then add the additional area in the hallway that was painted.
The wall that was painted was 9 feet by 10 feet, or 90 square feet. 90 square feet plus the additional 46 square feet in the hallway is 136 square feet, which is the total area that the paint covered.
Therefore the solution is 136 square feet.
Answer:
x=7
y=9
Step-by-step explanation:
x+7y=70 ...(1)
3x-6y=-33
divide by 3
x-2y=-11 ...(2)
(1)-(2) gives
x+7y-x+2y=70+11
9y=81
y=9
x-2*9=-11
or x=-11+18=7
Answer:
C
Step-by-step explanation:
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Yes, since 50 x 10 = 500.
