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:
=4x6−x3+x−3
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
3 rectangles are 3 x 7 x 2 = 42
2 triangles are 2 x 1/2 x 2 x 
= 2
note that you have to use Pythagoras Theorem to solve for the height of the triangle
use unitary method
for 5 it costs 330p
so for 1 choc it will cost 330/5
to get 9,
(330/5) =66
66 multiplied by 9 =594
therefore 9 will cost 594p
I think it would be y=1x+(0-2)
