<u>Answer:</u>
A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.
<u>Solution:</u>
We need to show that the gradient of the curve at A is 1
Here given that ,
--- equation 1
Also, according to question at point A (b+1,0)
So curve at point A will, put the value of x and y

0=b+1-c --- equation 2
According to multiple rule of Differentiation,

so, we get



By putting value of point A and putting value of eq 2 we get


Hence proved that the gradient of the curve at A is 1.
Answer:
not sure about 1
for number 2 it is 5 over 4 . for y, 5 10 15 20 is a multiple of 5 so the change is 5. for x, 4 8 12 16 is a multiple of 4 so the change is 4.
^ means to the power of or exponent
logb(x) = y is b^y = x
Answer:
x=24
Step-by-step explanation:
1/2x+2 = 1/2x^2+8
1/2x^2=1/4x
1/2x+2=1/4x+8
1/2x -1/4x =0.25x
0.25x+2=8
8-2=6
0.25x=6
0.25/6 = 24
x=24
Answer: D
Step-by-step explanation:
Because the Midrange data set shown is 55