1. Take an arbitrary point that lies on the first line y=3x+10. Let x=0, then y=10 and point has coordinates (0,10).
2. Use formula 
to find the distance from point 
to the line Ax+By+C=0.
The second line has equation y=3x-20, that is 3x-y-20=0. By the previous formula the distance from the point (0,10) to the line 3x-y-20=0 is:
.
3. Since lines y=3x+10 and y=3x-20 are parallel, then the distance between these lines are the same as the distance from an arbitrary point from the first line to the second line.
Answer: 
.
Answer:
D
Step-by-step explanation:
Since you are trying to make x on its own, you are trying to get rid of the co-efficient. The co-efficient is three, so you divide 3 by 3, which would equal one, leaving x on its own. Whatever you do to one side you do to the other, so you would divide 27 by 3 as well.
x = 9
33)31
34) 31
35) 21
Check the last one to make sure it’s right . The first 2 I’m 100% sure they are correct
Answer:
40.5 cm
Step-by-step explanation:
So think about it what is the Formula for a Trapezoid? Well it's h(b1+b2)/2! So we use this formula and place the numbers in there! When you're done you should have this: 6(9+4.5)/2. Now we will do <u>PEMDAS</u> (parenthesis, exponents, multiplication, division, addition, subtraction). from left to right. So we first do 9+4.5. That equals 13.5. Then we times that by 6, which is 81. Then we finally divide it by 2. Then there's your answer! :D
Answer:
9/32
Step-by-step explanation:
BC = black circle
BS = black square
WC = white circle
WS = white square
Given:
(WC + WS) / (BC + BS) = 5 / 11
WC / WS = 3 / 7
BC / BS = 3 / 8
Find: (BC + WC) / (BC + BS + WC + WS)
Solve for WC and BC in the last two equations, then substitute into the first:
WC = 3/7 WS
BC = 3/8 BS
WC + WS = 5/11 (BC + BS)
3/7 WS + WS = 5/11 (3/8 BS + BS)
10/7 WS = 5/8 BS
WS = 7/16 BS
Therefore:
WC = 3/7 WS
WC = 3/16 BS
Substitute:
(BC + WC) / (BC + BS + WC + WS)
(3/8 BS + 3/16 BS) / (3/8 BS + BS + 3/16 BS + 7/16 BS)
(9/16 BS) / (2 BS)
9/32
