The diagonal of a rectangle = sqrt(w^2 + l^2)
w = width
l = length
In this problem,
The diagonal = 20 in
w = x
l = 2x + 8
Let's plug our numbers into the formula above.
20in = sqrt((x)^2 + (2x + 8)^2)
Let's simplify the inside of the sqrt
20 in = sqrt(5x^2 + 32x + 64)
Now, let's square both sides.
400 = 5x^2 + 32x + 64
Subtract 400 from both sides.
0 = 5x^2 + 32x - 336
Factor
0 = (5x - 28)(x + 12)
Set both terms equal to zero and solve.
x + 12 = 0
Subtract 12 from both sides.
x = -12
5x - 28 = 0
Add 28 to both sides.
5x = 28
Divide both sides by 5
x = 28/5
The width cant be a negative number so now we know that the only real solution is 28/5
Let's plug 28/5 into our length equation.
Length = 2(28/5) + 8 = 56/5 + 8 = 96/5
In conclusion,
Length = 96/5 inches
Width = 28/5
Answer:
Y>-4
Step-by-step explanation:
Add 11 to both sides and divide
Hello, please consider the following.
We will multiply the numerator and denominator by
to get rid of the root in the denominator.
First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.
We need to take any x real number different from 8/3 then and simplify the expression.
Let's do it!
Thank you
