Answer:
Math is a continuous function. It can keep going on and on.
Step-by-step explanation:
Hope I <em><u>Helped!</u></em> :D
RemarkIf you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step OneDivide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step TwoDifferentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =

Frankly I like the first answer better, but you have a choice of both.
Okay, so a general rule for finding perpendicular lines in the form of y = mx + b is y = (-1/m) + b.
First, let's ignore b (-7) because we're going to find that later.
A perpendicular line to y = 4x + b is y = -1/4x + b.
Alright, so now let's plug in the values. They are in the form of (x,y), so let's plug them in accordingly.
3 = -1/4(4) + b
3 = -1 + b
b = 4
y = -1/4x + 4
So a line perpendicular to y = 4x - 7 is y = -1/4x + 4.
This can be solved using a ratio. The ratio is: 13/20=x/60 this would then become 20x= 780 divide both sides by 20 and you get x=39. the answer is 39