Answer:
if I'm not wrong I think it's B
Step-by-step explanation:
step 1. I guess we assumed the two lines across the transversal are parallel
step 2. 3x + 1 = 85 (definition of alternate exterior angles)
step 3. 3x = 84 (subtract 1 from each side)
step 4. x = 28. (divide both sides by 3)
<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.
Expand the expression?
Perform indicated multiplication:
7a(2a)+7a(-5)-4(2a)-4(-5)
14a^2-35a-8a+20 combine like terms, ie (-35a-8a)=-43a
14a^2-43a+20