The solutions to this equasion are x = 0 and y=1.
Step-by-step explanation:
GIVEN,
f(x)=2x+1; g(x)=3x-2
NOW,
( 2x+1 + 3x-2)×3
=(5x-2)×3
=15x - 6 <em>A</em><em>N</em><em>S</em><em>W</em><em>E</em><em>R</em>
Answer:
You would use a^2 + b^2 = c^2
Step-by-step explanation:
a and b are the shorter side of the triangle whereas c is the hypotenuse or longest side of the triangle
Answer:
Domain = {All real values of x EXCEPT x = -5 and x = 7}
Step-by-step explanation:
This is a rational function given as y=\frac{6+9x}{6-|x-1|}y=
6−∣x−1∣
6+9x
The domain is the set of all real value of x for which the function is defined.
For rational functions, we need to find which value of x makes the denominator equal to 0. We need to exclude those values from the domain.
Now
6 - |x-1| = 0
|x-1| = 6
x- 1 = 6
or
-(x-1) = 6
x = 6+1 = 7
and
-x+1=6
x = 1-6 = -5
So, the x values of -5 and 7 makes this function undefined. So the domain is the set of all real numbers except x = -5 and x = 7
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.