Pete has an old playstation that he bought for $400.
2 answers:
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Answer:
Option B.
Step-by-step explanation:
Consider the below figure attached with this question.
It is given that triangle ABC is an isosceles triangle.
From the below figure it is clear that the vertices of triangle are A(-2,-4), B(2,-1) and C(3,-4).
Distance formula:
Using this formula, we get
Now,
Perimeter of triangle ABC = AB + BC + AC
So, perimeter of triangle ABC is units.
Therefore, the correct option is B.
Answer:
For a rectangular solid with:
width = w
length = l
height = h
The volume is equal to:
V = w*l*h
in this case we know that:
width = w.
"length of 7 more than the width"
l = w + 7.
"a height that is equivalent to 15 decreased by 3 times the width."
h = 15 - 3*w
Then the volume will be:
V = w*l*h = w*(w + 7)*(15 - 3w) = (w^2 + 7*w)*(15 - 3*w)
V = ( -3*w^3 + 15*w^2 + 105*w - 21*w^2)
V = (-3*w^3 - 6*w^2 + 105*w)
Now, the maximum volume will be for the value of w such that:
V'(w) = 0.
and:
V''(W) < 0
Where:
dV/dw = V'(w).
dV'/dw = V''(w)
Then first we need to differentiate the equation for the volume.
V'(w) = dV/dw = ( 3*(-3*w^2) + 2*(-6*w) + 105)
V'(w) = -9*w^2 - 12*w + 105.
Then we need to find the solution for:
-9*w^2 - 12*w + 105 = 0.
We can use the Bhaskara formula, and we will get:
Then the two solutions are:
w₁ = (+12 - 62.6)/(-18) = 2.81
w₂ = (+12 + 62.6)/(-18) = -15.5
But we can not have a negative width, so we can just discard the second solution.
Now let's check the second condition for the maximum, we must have:
V''(2.81) < 0.
V'' = dV'/dw = 2*(-9*w) - 12 = -18*w - 12
V''(2.81) = -18*2.81 - 12 = -62.58 < 0 .
Then the volume is maximized when w = 2.81, and the maximum volume will be:
V(2.81) = (-3*(2.81)^3 - 6*(2.81)^2 + 105*2.81) = 180.1