The area of D is given by:
The average value of f over D is given by:
![\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B343%7D%7B3%7D%20%7D%20%20%5Cint%5Climits%5E7_0%20%20%5Cint%5Climits%5E%7Bx%5E2%7D_0%20%7B4x%5Csin%28y%29%7D%20%5C%2C%20dydx%20%20%3D%20-%5Cfrac%7B3%7D%7B343%7D%20%20%5Cint%5Climits%5E7_0%20%7B4x%5Ccos%28x%5E2%29%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D-%5Cfrac%7B3%7D%7B343%7D%20%5Cint%5Climits%5E%7B49%7D_0%20%7B2%5Ccos%28t%29%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%20%5Cleft%5B%5Csin%28t%29%5Cright%5D_0%5E%7B49%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%5Csin49)
So you know the compound continuously formula is
. Thus, just plug in.
Plug that into your calc to get $1832.
We have that
<span>y=2x+4--------> equation 1
3x−6y=3-------> equation 2
step 1
</span>I substitute the value of y in equation 1 for the value of y in equation 2<span>
so
</span>3x−6*[2x+4]=3-------> 3x-12x-24=3
-9x=3+24
-9x=27------> 9x=-27
x=-27/9
x=-3
step 2
<span>I substitute the value of x in equation 1 to get the value of y</span>
y=2x+4--------> y=2*(-3)+4--------> y=-6+4
y=-2
the answer is
the solution is the point (-3,-2)
x=-3
y=-2
From the description given for the triangle above, I think the type of triangle that is represented would be a right triangle. This type of triangle contains a right angle and two acute angles. In order to say or prove that it is a right triangle, it should be able to satisfy the Pythagorean Theorem which relates the sides of the triangle. It is expressed as follows:
c^2 = a^2 + b^2
where c is the hypotenuse or the longest side and a, b are the two shorter sides.
To prove that the triangle is indeed a right triangle, we use the equation above.
c^2 = a^2 + b^2
c^2 = 20^2 = 10^2 + (10sqrt(3))^2
400 = 100 + (100(3))
400 = 400
