Solve each equation separately:
(-1, 3) Has to be a point that fits both equations
y = 2x + ____
Plug in Values:
3 = 2(-1) + ____
3 = -2 + ____
___ = 5
First Blank: 5
y = ___x - 1
Plug in Values:
3 = ___(-1) - 1
4 = ___(-1)
___ = -4
Second Blank: -4
Answer:
1. attachment↑
its grade 8
Step-by-step explanation:
mark me as brainliest if it helps
Answer:
12
Step-by-step explanation:
you have to times 2 x 6 to find your x
I'm not for sure tho but I hope it's right
Answer:
At (-2,0) gradient is -4 ; At (2,0) gradient is 4
Step-by-step explanation:
For this problem, we simply need to take the derivative of the function and evaluate when y = 0 (when crossing the x-axis).
y = x^2 - 4
y' = 2x
The function y = x^2 - 4 cross the x-axis when:
y = x^2 - 4
0 = x^2 - 4
4 = x^2
2 +/- = x
Hence, this curve crosses the x-axis twice, once at (-2,0) and again at (2,0).
The gradient at these points are as follows:
y' = 2(-2) = -4
y' = 2(2) = 4
Cheers.
Answer:
384
Step-by-step explanation:
64*6=384
Hope it helped!
