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
We see that they are made up of identical squares, meaning that each side of each square is the same length. There are 12 square sides that make up the perimeter of A, so we can do 72/12 which is 6. We see there are ten total square lengths on the perimeter of b, so we can do 6 times 10 which is 60 cm. That’s your perimeter.
Answer:
The y-coordinate of the solution is -5.
Step-by-step explanation:
I graphed the equations on the graph below to find the solution of the system.
If this answer is correct, please make me Brainliest!
To find the surface area, multiply length x width x height.
let "length" = l
l x 32 x 36 = 4344
Simplify
l x (32 x 36) = 4344
l x 1152 = 4344
Isolate the length. Divide 1152 from both sides
(l x 1152)/1152 = (4344)/1152
l = 4344/1152
l = 3.77 (rounded)
3.77 cm is your length.
hope this helps
Answer:
The finish line
Step-by-step explanation:
Jonas is planning his trip using the finish line as reference point. He is using an interval scale (using meters as measurement) and the finish line = 0 meters or the starting point. Anything before the finish is negative and anything after is positive. In this case, his wife is located at 91 + 14 = 105 meters away from him. When you use an interval scale to measure distance, you must use absolute values to determine the distance because positive and negative points.
