Answer:
1 $16.50
2 $28.50
3 $40.50
$4.50 + $12c
$100.50
Step-by-step explanation:
Please find attached the complete question
Total cost in dollars = fixed cost + (variable cost x number of CDs bought)
fixed cost = $4.50
variable cost = $12
number of cds bought = c
total cost in dollars = $4.50 + $12c
total cost in dollars when 1 cd is bought = $4.50 + $12(1) = $16.50
total cost in dollars when 2 cds are bought = $4.50 + $12(2) = $28.50
total cost in dollars when 3 cds are bought = $4.50 + $12(3) = $40.50
total cost in dollars when 8 cds are bought = $4.50 + $12(8) = $100.50
Answer:
(5x^2-3) +(2x^2-3x^3)
=3x^3+5x^2+2x^2-3
= 3x^3 +7x^2 -3
Step-by-step explanation:
hope this is correct cuz I'm not an expert...
Answer:
- increasing: (π/2, 3π/2)
- decreasing: [0, π/2) ∪ (3π/2, 2π]
- minimum: -16 at x=π/2
- maximum: 16 at x=3π/2
Step-by-step explanation:
If all you want are answers to the questions, a graphing calculator can provide them quickly and easily. (see attached)
___
If you need an algebraic solution, you need to find the zeros of the derivative.
f'(x) = -16cos(x)sin(x) -16cos(x) = -16cos(x)(sin(x) +1)
The product is zero where the factors are zero, at x=π/2 and x=3π/2.
These are the turning points, where the function changes from decreasing to increasing and vice versa.
(sin(x)+1) is non-negative everywhere, so the sign of the derivative is the opposite of the sign of the cosine function. This tells us the function f(x) is increasing on the interval (π/2, 3π/2), and decreasing elsewhere (except where the derivative is zero).
The function local extrema will be where the derivative is zero, so at f(π/2) (minimum) and f(3π/2) (maximum). We already know that cos(x) is zero there, so the extremes match those of -16sin(x).
To find the amount of fencing needed, we will find tge perimeter
To find the perimeter;
Find the distance |AB|, |BC|, |CD| and |DA|
We will use the formula below to find the distances:
![|d|=\sqrt[]{(x_2-x_1)^2+(y_{2-}y_1)^2}](https://tex.z-dn.net/?f=%7Cd%7C%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_%7B2-%7Dy_1%29%5E2%7D)
Distance |AB|
A(7,7), B(16,7)
x₁=7 y₁=7 x₂=16 y₂=7
Substituting into the formula;
![|AB|=\sqrt[]{(16-7)^2+(7-7)^2}](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%5B%5D%7B%2816-7%29%5E2%2B%287-7%29%5E2%7D)
![=\sqrt[]{9^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B9%5E2%2B0%7D)
![=\sqrt[]{9}^2\text{ =9}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B9%7D%5E2%5Ctext%7B%20%3D9%7D)
Distance |BC|
B(16,7), C(2,2)
x₁=16 y₁=7 x₂=2 y₂=2
substituting into the formula;
![|BC|=\sqrt[]{(2-16)^2+(2-7)^2}](https://tex.z-dn.net/?f=%7CBC%7C%3D%5Csqrt%5B%5D%7B%282-16%29%5E2%2B%282-7%29%5E2%7D)
![=\sqrt[]{(-14)^2+(5)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-14%29%5E2%2B%285%29%5E2%7D)
![=\sqrt[]{196+25}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B196%2B25%7D)
![=\sqrt[]{221}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B221%7D)
Distance |CD|
C(2,2), D(16,2)
x₁=2 y₁=2 x₂=16 y₂=2
substituting into the formula;
![|CD|=\sqrt[]{(16-2)^2+(2-2)^2}](https://tex.z-dn.net/?f=%7CCD%7C%3D%5Csqrt%5B%5D%7B%2816-2%29%5E2%2B%282-2%29%5E2%7D)
![=\sqrt[]{14^2+0}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%2B0%7D)
![=\sqrt[]{14^2}=14](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B14%5E2%7D%3D14)
Distance |DA|
D(16,2) A(7,7)
x₁=16 y₁=2 x₂=7 y₂=7
Substituting into the formula;
![|DA|=\sqrt[]{(7-16)^2+(7-2)^2}](https://tex.z-dn.net/?f=%7CDA%7C%3D%5Csqrt%5B%5D%7B%287-16%29%5E2%2B%287-2%29%5E2%7D)
![=\sqrt[]{(-9)^2+(5)^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B%28-9%29%5E2%2B%285%29%5E2%7D)
![=\sqrt[]{81+25}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B81%2B25%7D)
![=\sqrt[]{106}=10.30](https://tex.z-dn.net/?f=%3D%5Csqrt%5B%5D%7B106%7D%3D10.30)
Perimeter = |AB|+|BC|+|CD|+|DA|
= 9 + 14.87 + 14 + 10.30
=48.17
≈48
Hence;
48 feet of fencing is needed
Answer:
I think it's 48 but I'm not sure you'll have to wait and see if a smarter person than me knows it, sorry.
