Answer:
D. ![x^2 + 3x - 88](https://tex.z-dn.net/?f=%20x%5E2%20%2B%203x%20-%2088%20)
Step-by-step explanation:
Area of rectangle = ![length (l) * width (w)](https://tex.z-dn.net/?f=%20length%20%28l%29%20%2A%20width%20%28w%29%20)
Length of rectangle = (x - 8)
Width = (x + 11)
Area of rectangle = ![(x - 8)(x + 11)](https://tex.z-dn.net/?f=%20%28x%20-%208%29%28x%20%2B%2011%29%20)
Expand the expression
(distributive property of multiplication)
![x^2 + 11x - 8x - 88](https://tex.z-dn.net/?f=%20x%5E2%20%2B%2011x%20-%208x%20-%2088%20)
Combine like terms
![x^2 + 3x - 88](https://tex.z-dn.net/?f=%20x%5E2%20%2B%203x%20-%2088%20)
Expression for the area of the rectangle = ![x^2 + 3x - 88](https://tex.z-dn.net/?f=%20x%5E2%20%2B%203x%20-%2088%20)
Answer:
50
Step-by-step explanation:
There is 180 degrees in a triangle.
65 + 65 = 130
180 - 130 = 50
Answer:
Step-by-step explanation:
A rectangle is shown with length x plus 10 and width 2 x plus 5.
so the rectangle area = length * width
= (x + 10) * (2x + 5)
an unshaded square with length x plus 1 and width x plus 1
so the unshaded square area = (x + 1) * (x + 1)
the shaded area = rectangle area - square area
= (x + 10) * (2x + 5) - (x + 1) * (x + 1)
= (2x^2 + 20x + 5x + 50) - (x^2 + x + x + 1)
= x^2 + 23x + 49
Copper: Nickel = 4:1
a) How much copper needed for 20 kg nickel
C:N = 4: 1
C:20 = 4:1
C/20 = 4/1 Cross Multiply
C*1 = 4*20
C = 80
80kg Copper would be needed.
b) How much Nickel would be needed for 20 kg of Copper
C:N = 4: 1
20:N = 4:1
20/N = 4/1 Cross Multiply
4*N = 1*20
4N = 20 Divide both sides by N.
N = 20/4 = 5
5kg Nickel would be needed.
Answer:
0.42
Step-by-step explanation:
use the attached as reference
recall that for a right (90°) triangle, with any acute angle θ,
cos θ = length of adjacent side / length of hypotenuse
in our case, for angle I, the adjacent side is HI = 36 units and the hypotenuse is IJ = 85 units
by the above formula,
cos I = length of adjacent side / length of hypotenuse
= 36 / 85
= 0.42353
= 0.42 (rounded to nearest hundredth)