Suppose

has two roots, reciprocals of one another. Call them

, with

.
Let's divide through

by

for now. By the fundamental theorem of algebra, we can factorize

as

Expand the RHS to get

so we must have


The first equation says

and

occur in a ratio of the negative sum of the roots of

, while the third equation says that the first and last coefficients

must be the same.
The answer is The median is 69
Answer:
The power developed by the 1st crane is twice the power developed by the 2nd crane.
Step-by-step explanation:
According to the question,
work done by both of the cranes are the same. Let it be W.
Then power developed by the first crane,
=
unit -----------------(1)
and, the power developed by the 2nd crane,
=
unit ------------------(2)
So, we get that,

So, the power developed by the 1st crane is twice the power developed by the 2nd crane.
Answer:
Step-by-step explanation:
We can see that this graph looks something like the graph of:
f(x) = 1/x^2
But the asymptotes of 1/x^2 are at x = 0, and in this case we can see that the asymptotes are near x = -3.
This may mean that the graph has been horizontally shifted 3 units to the left.
A general way to write an horizontal shift of N units is:
g(x) = f(x + N)
if N is positive, then the shift is to the left, if N is negative, then the shift is to the right.
In this case, we will have;
g(x) = f(x + 3)
And f(x) = 1/x^2
then:
g(x) = 1/(x + 3)^2
Graphing that, we get the graph shown below, that is almost the same as the graph in the image.