One side of a rectangular yard is bounded by the side of a house. The other three sides are to be fenced with 425 ft of fencing. The length of fence opposite the house is 10 ft less than either of the other two sides. Find the dimensions (in ft) of the yard.

shorter side ft

longer side ft

Longer side = 145ft

Shorter side = 135ft

Step-by-step explanation:

Here we have a rectangle.

Let's define L as the longest side (the one that is perpendicular to the wall of the house) and S as the shorter side.

We know that the perimeter of a rectangle is:

P = 2*L + 2*S

So if we wanted to fence this rectangle, we would need fencing equal to the perimeter.

But in this case, one of the shorter sides does not need to be fenced, so we only need:

2*L + S of fencing.

And we know that we will use 425ft of fencing, then:

2*L + S = 425 ft

We also know that The length of the fence opposite the house is 10 ft less than  either of the other two sides, then:

S = L - 10ft

Then we have two equations:

2*L + S = 425 ft

S = L - 10ft

Here we can just replace the second equation into the first one to get:

2*L + (L - 10ft) = 425 ft

Now we can solve this for L

2*L + L - 10ft = 425ft

3*L = 425ft + 10ft

3*L = 435ft

L = 435ft/3 = 145ft

This means that the longer side is 145ft long.

And by the equation:

S = L - 10ft

S = 145ft - 10ft = 135ft

S = 135ft

We know that the shorter side is 135ft long.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length