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
Answer: I would say yes because real numbers include rational numbers, integers, whole, Natural, and Irrational.
* Hopefully this helps:)!! Mark me the brainliest:)!!
* Hopefully the picture also helps because it really helped me.
<em>∞ 234483279c20∞</em>
Best is to draw a sketch of the three points.
Next step is to find the distances BC, CD, DB.
The perimeter is the sum of the three distances.
The distances are found using the distance formula:
D=sqrt((y2-y1)^2+(x2-x1)^2)
order of (x1,y1), (x2,y2) is not important.
BC=sqrt((3- -3)^2+(5-3)^2)=sqrt(6^2+2^2)=sqrt(40)=5.324
CD=sqrt((-1-3)^2+(0-5)^2)=sqrt(4^2+5^2)=sqrt(41)=6.403
DB=sqrt((-3- -1)^2+(3-0)^2)=sqrt(2^2+3^2)=sqrt(13)=3.606
So perimeter = BC+CD+DB=16.333 (approximately)
Just multiply each times 0.15 :)
$35.15 = $5.2725
&27.00 = $4.05
I think it’s either B or D
