OK, so she definitely drove home faster than when she drove to her first destination. So she must've taken quicker to get home. Lets assume 3 hours on the way there and 2 hours on the way back.
I really hoped this helped. I'm no genius, but you asked for help and I tried to give it.
Have a great day.
Answer:
y+2x=35
so how to get that answer if you divided 35/2= 17.5 so 17 *2= 34 +y= 35
y =1
round it down one number so y can equal 1 then you get 35
Your Answer: G(x) = (x − 5)²
I just finished my semester exam, and this was one of the questions. I guessed on this question because I had no clue of what the answer was, and lucky me, I got the question right :D
Hope this helps y'all with your test :)
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
Answer:
m=-2
Step-by-step explanation:
We will use the formula for rate of change (slope).
m=(y2-y1)/(x2-x1)
Pick any pair of x and y values.
(10-12)/(2-1)
Simplify.
m=-2
HTH :)