Answer:
19
----- = x
40-3a
Step-by-step explanation:
3(ax + 9) = -4 (-2 - 10x)
Distribute
3ax +27 = 8+40x
Subtract 3ax from each side
3ax-3ax +27 = 8+40x-3ax
27 = = 8+40x-3ax
Subtract 8 from each side
27-8 = 8-8+40x-3ax
19 = 40x-3ax
Factor an x on the right side
19 = x(40-3a)
Divide each side by 40-3a
19/(40-3a) = x(40-3a)/(40-3a)
19
----- = x
40-3a
The point of elimination is to have one variable so example
3x + 2y = 4
7X + 3y = 5
1) Try to find a way to have one variable
7(3x + 2y = 4) ----> 21x + 14y = 28
-3(7x+ 3y = 5) -----> -21x -9y = -15
2) add
0 + 5y = 13
3) solve for the last variable
5y = 13
y = 13/5 = 2 3/5 = 2.8
4) Subsitute the variable to get the other one
3x + 2(2.8) = 4
3x + 5.6 = 4
3x = -1.6
x = 0.533333333 (continue)
The intersection would be 0.53 with a line above the 3 and 2.8. (0.5333 , 2.8)
Hope you understand!!
Answer:
5
Step-by-step explanation:
5*20=100
Area of a rectangle:
A = w x h, where w is the width and h is the height.
w = 2.25 yd
A = 1.5 yd²
1.5 = 2.25 · h
h = 1.5 : 2.25 = 0.6667 = 2/3 yd
Answer:
The height of a banner is 2/3 yard.
Area of the rectangular mat = 12 x 4 = 48 in²
Area of the 3 circles = 3( πr²) = 3(3.14 x 2²) = 37.68 in²
P(landing in a circle) = 37.68/48 = 157/200 = 0.79 (nearest hundredth)
Answer: The probability of landing in one of the circles is 0.79.
