First we find the zeroes so we don't take the integral of negative bits
4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4
![\int\limits^4_0 {4x-x^2} \, dx =[2x^2- \frac{1}{3}x^3]^4_0=(32- \frac{64}{3})-(0)= 10.6666666666](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E4_0%20%7B4x-x%5E2%7D%20%5C%2C%20dx%20%3D%5B2x%5E2-%20%5Cfrac%7B1%7D%7B3%7Dx%5E3%5D%5E4_0%3D%2832-%20%5Cfrac%7B64%7D%7B3%7D%29-%280%29%3D%20%2010.6666666666)
or aout 10 and 2/3
C is answer
I think the answer is like 0.15¢ assuming the way to solve it is to divide the cost by the Oz per pkg which is 48 so 6.99/48 is 0.15¢ rounded
Answer:
A = $2,240.00
(I = A - P = $240.00)
Equation:
A = P(1 + rt)
Calculation:
First, converting R percent to r a decimal
r = R/100 = 3%/100 = 0.03 per year.
Solving our equation:
A = 2000(1 + (0.03 × 4)) = 2240
A = $2,240.00
The total amount accrued, principal plus interest, from simple interest on a principal of $2,000.00 at a rate of 3% per year for 4 years is $2,240.00.
I hope i’m correct. Hope this helps!
Answer:
Infinite solutions
Step-by-step explanation:
Distribute the numbers
3(6x-2) = 18x - 6
2(9x-3) = 18x - 6
18x - 6 = 18 - 6
Since both sides of the equal sign are the same, that means you can plug any number into x, and it will always be a valid answer.
