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
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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.
Answer:
That equation is (2.5×2). The mean of the K is ×2
You add 2x and 7x to get 9x
so the answer is 9x+5y
Answer:
So for the first one, circle Ignored it becaue you cna't buy any more.
For number 2, Quotient is 8 & R is 10. Circle Rounded the quotient up because you need another bin to hold it.
Step-by-step explanation:
