I think it would be be because you’re actually saving a little bit more amount of money you’re paying a dollar and would say two cent for each bar
If you had values instead of algebraic expressions, you would find the area of the patio and subtract that from the area of the yard. Even though you don't have values, you can still find the area and subtract by expanding and simplifying:
(8x + 2)(6x + 3)
8x × 6x = 48x²
8x × 3 = 24x
2 × 6x = 12x
2 × 3 = 6
So you get 48x² + 36x + 6 as your area of the yard
(x + 5)(3x + 1)
x × 3x = 3x²
x × 1 = x
5 × 3x = 15x
5 × 1 = 5
So the area of the patio is 3x² + 16x + 5
(48x² + 36x + 6) - (3x² + 16x + 5)
48x² - 3x² = 45x²
36x - 16x = 20x
6 - 5 = 1
So your answer is D. 45x² + 20x + 1. I hope this helps!
Answer:
1999
Step-by-step explanation:
y = 0.2313(y - 1992)^2 + 2.600(y - 1992) + 35.17
verify with known data point
y = 0.2313(2004 - 1992)^2 + 2.600(2004 - 1992) + 35.17
y = 0.2313(12)^2 + 2.600(12) + 35.17
y = 27.9873 + 28.6 + 35.17
y = 99.6772 which verifies our equation
65 = 0.2313x² + 2.6x + 35.17
0 = 0.2313x² + 2.6x - 29.83
quadratic formula
x = (-2.6 ±√(2.6² - 4(0.2313)(-29.83))) / (2(0.2313))
x = (-2.6 + 5.86) / 0.4626 = 7.05 years
7.05 = y - 1992
y = 1999.05
Answer:
iits F(x)+440-201
Step-by-step explanation:
