Answer:
D. 10
Step-by-step explanation:
To find the slope of the equation, we first need to put the equation into slope-intercept form:
y=mx+b
So first, we need to get y by itself.
10x-y=2
Subtract 10x from both sides.
-y=-10x+2
Then, divide -1 on both sides.
y=10x-2
Slope=10
The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
_____
You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
(33)(1/9•2-6+2)
18-6+2=14
33•1/14 = 33/14
The answer is 33/14 or 2.35714
It would be 31+t.
Unless there's more to this question.
