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)
If 1/5 can for 1 (each) room, then:
for 51 rooms, (51*1/5)=10.2 rooms.
Round to nearest whole number since you can't just paint 0.2 of a room: answer = 10
Answer:
1/2
Step-by-step explanation:
1/4=2/8
2/8+2/8=4/8=1/2
Answer:
Quadrant 1
Step-by-step explanation:
how I understood this is first understanding how a graph works. Once you understand that you can count 13 horizontal on the "X" Axis, then go 18 up on the "Y" axis. you should get a point in the upper left area and that area is called Quadrant 1. Say if 13 was negative instead of going horizontal east you would go horizontal west. And the point should end up in Quadrant 2. Any more questions feel free to ask me below.
Answer:
The correct order is:
a
c
d
b
Step-by-step explanation:
First, let's write 1/x in a convenient way for us:
a) Substitute 1/x = p/q, to obtain x = 1/(1/x) = 1/(p/q) = q/p.
Now we assume that 1/x is rational (we want to prove that this implies that x will be also rational and because we know that x is irrational assuming that 1/x is rational will lead to an incongruence), then:
c. If 1/x is rational, then 1/x = p/q for some integers p and q with q ≠ 0. Observe that p is not 0 either, because 1/x is not 0.
Now we know that we can write x as a quotient of two integers, we need to imply that, then the next one is:
d) Observe that x is the quotient of two integers with the denominator nonzero.
And that is the definition of rational, then we end with:
b) Hence x is rational.
Which is what we wanted to get.
