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:
109
Step-by-step explanation:
AB = 180° ( straight line )
X = 180 - 42 - 29
= 138 - 29
= 109
3 is 25% of 12, so if you apply the same proportion to 180, a quarter of 180 is 45!!
Answer:
The first 4 are correct and the last one is incorrect
Step-by-step explanation:
1. The diameter of the circle is 30.8 m ( Correct )
Radius = 15.4 so diameter is 2 times the radius which is 30.8 m
2. The circumference in terms of π is 30.8 π ( Correct )
Circumference = π × Diameter
Circumference = π × 30.8
Circumference = 30.8 π
3 . The circumference of the circle can be found using 2 × π × 15 . 4 ( Correct )
Circumference = π × Diameter
Circumference = π × 30.8
Circumference = 30.8 π
4 . The approximate circumference of the circle rounded to the nearest tenth is 96.7 m ( Correct )
Circumference = 30.8 π = 96.7610537306
5 . A little more than 6 diameters could be wrapped around the circle
( False )
Circumference = 96.7610537306
Diameter = 30 . 8
96.7610537306 ÷ 30 . 8 = 3.14
