Answer:
Step-by-step explanation:
This equation has no solution where 'a' and 'b'
are both non-zero real numbers.
You said 2a + 2b = 2a
Subtract 2a from each side: 2b = 0
'b' must be zero.
And it doesn't matter what the value of 'a' is.
As long as b=0, 'a' can be any number.
Answer:
Suppose that A and B are points on the number line.
If AB=11 and A lies at 6, where could B be located?
If there is more than one location, separate them with commas
Step-by-step explanation:
Suppose that A and B are points on the number line.
If AB=11 and A lies at 6, where could B be located?
If there is more than one location, separate them with commas
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.
