Given:
The equation of the curve is:

To find:
The gradient (slope) of the given curve at point (2,7).
Solution:
We have,

Differentiate the given equation with respect to x.


Now we need to find the value of this derivative at (2,7).




Therefore, the gradient (slope) of the given curve at point (2,7) is 19.
The gcf of 64 and 16 is16
Well, first of all, the first statement (ABC = ADC) looks like it just says
that the two halves of the little square ... each side of the diagonal ...
are congruent. That's no big deal, and it's no help in answering the
question.
The effect of the dilation is that all the DIMENSIONS of the square
are doubled ... each side of the square becomes twice as long.
Then, when you multiply (length x width) to get the area, you'd have
Area = (2 x original length) x (2 x original width)
and that's
the same as (2 x 2) x (original length x original width)
= (4) x (original area) .
Here's an easy, useful factoid to memorize:
-- Dilate a line (1 dimension) by 'x' times . . . multiply the length by x¹
-- Dilate a shape (2 dimensions) by 'x' . . . multiply area by x²
-- Dilate a solid (3 dimensions) by 'x' . . . multiply volume by x³
And that's all the dimensions we have in our world.
_______________________________
Oh, BTW . . .
-- Dilate a point (0 dimensions) by 'x' . . . multiply it by x⁰ (1)