Area of rectangle = length × width
300 = (x+10) × x
300 = 10x + x^2
Now we have the equation y=x^2 + 10x - 300
We can factor this equation to get our zeroes, which are:
-5(1+sqrt(13)),
5(sqrt(13)-1)
We only use the positive zero because we can't have a negative dimension.
Your width would be:
5(sqrt(13)-1)
Your length would be:
5(sqrt(13)-1) + 10
Answer:
bbbbbbbbbbbbbbbbbbbbbbbbb
Answer:
Step-by-step explanation:
F(x) results in a parabola with vertex (0,0) wich mean there is only one x-int at that point. g(x) has been shifted the grapgh of f(x) to the right by to units and down by three unites. Now our vertex lies in the point (2,-3) and since the graph was move dow i=of the x-axis we now have two different x-intercepts.
Answer:
Before you get started, take this readiness quiz.
Write as an inequality: x is at least 30.
If you missed this problem, review (Figure).
Solve 8-3y<41.
If you missed this problem, review
the function is given, and it's value is where the object is ("how far to the right").
so as long as it rises (going more right), this will be apply.
in the screenshot I graphed the function. of course t is graphed as x and "along the x-axis" is graphed as y, but the pattern is the same anyways.
for the first 1.25 seconds the object goes to the left, and after that always to the right.
since we look at t to calculate x, t effectively takes the role of the important variable that is normally given to x. the calculation pattern are just the same. so let's find the lowest point of this function by calculating it out.
x(t) = 2t² – 5t – 18
x'(t) = 4t -5
x'(t) = 0
0 = 4t -5
5 = 4t
1.25 = t
plugging it into the second derivative
x''(t) = 4
x''(1.25) = 4
it's positive, so at t=1.25 there is a low point
(of course the second derivative is constant anyways.)
the object is traveling toward the right
the object is traveling toward the rightfor t > 1.25