Triangle JKL has a base that is 1 unit long
Triangle J'K'L' has a base that is 3 units long
3 units / 1 unit = 3
The scale factor is 3.
The equation of the vertical line that satisfies the condition is x=0.
I think it is 90
Hope this help you?!
Answer:
14. x=-2.5, y = -7
15. x=28 y = -20
Step-by-step explanation:
14. Let's solve this system by elimination. Multiply the first equation by -1.
-1*(4x-y)= -1*(- 3)
-4x +y = 3
Then add this to the second equation.
-4x+y = 3
6x-y= - 8
--------------
2x = -5
Divide each side by 2
x = -2.5
We still need to find y
-4x+y =3
-4(-2.5) + y =3
10 +y =3
Subtract 10 from each side.
y = 3-10
y = -7
15.I will again use elimination to solve this system, because using substitution will give me fractions which are harder to work with. I will elimiate the y variable. Multiply the first equation by 11
11(
5x+6y)= 11*20
55x+66y = 220
Multiply the second equation by -6
-6(9x+11y)=32*(-6)
-54x-66y = -192
Add the modified equations together.
55x+66y = 220
-54x-66y = -192
---------------------------
x = 28
We still need to solve for y
5x+6y = 20
5*28 + 6y =20
140 + 6y = 20
Subtract 140 from each side
6y = -120
Divide by 6
y = -20
