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.
Let's start with triangle RST. This is a 30-60-90 triangle, which means it has the relationship x - x sqrt(3) - 2x.
If RS is 2 sqrt(3), then ST must be 2 and RT must be 4.
Triangle QRT is a 45-45-90 triangle, which means it has the relationship x - x - x sqrt(2).
If RT is 4, then RQ must also be 4.
Answer: x = 4
Hope this helps!
I would go with B i had this question on my test and i got it correct
You plug in the values so it would be 24=15k. Divide by 15 on both sides so k=24/15=1.6. :)
