Here you're being asked to find the "perimeter" of the space, even tho' the problem doesn't specifically ask for it.
The formula for P is P = 2W + 2L.
Here the width, W, is 3 1/2 yds, and the length, L, is 4 2/3 yds. Subbing these two values into the formula for P (above) results in:
P = 2(3 1/2 yds) + 2(4 2/3 yds)
= 7 yds + 9 1/3 yds = 16 1/3 yds, total.
Square root is this 4 and -4 are square roots to 16 because 4^2= 16
Answer:
Which grade's question is this?
For this case we have the following type of equations:
Quadratic equation:
Linear equation:
We observe that when equating the equations we have:
Rewriting we have:
We obtain a polynomial of second degree, therefore, the maximum number of solutions that we can obtain is 2.
Answer:
The greatest number of possible solutions to this system is:
c.2
You solve for y and get
y=3/4x-5
Slope is 3/4
Y intercept is -5