nswer:
209.8
Step-by-step explanation:
the bottom is = 11 * 5
two of the sides are 11 * 9.3 / 2 because they are triangles, and then times 2 because there are two of them
the other two sides are 10.5 * 5 / 2 , and the * 2 because there are two of them again.
(11 * 5) + (11 * 9.3 * 2 / 2) + (10.5 * 5 * 2 / 2) =209.8
18,267, 2728,2729110,72829183836,261729340
Answer:
Required total charge is
coulombs per square meter.
Step-by-step explanation:
Given electric charge is dristributed over the disk,
so that the charge density at (x,y) is,

To find total charge on the disk let Q be the total charge and
so that,
where A is the surface of disk.



![=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%7D%7B3%7D%5Cint_%7B0%7D%5E%7B2%5Cpi%7D%28%5Csin%5Ctheta%2B%5Ccos%5Ctheta%29%5CBig%5Br%5E3%5CBig%5D_%7B0%7D%5E%7B4%7Dd%5Ctheta%2B2%5Cint_%7B0%7D%5E%7B2%5Cpi%7D%5CBig%5B%5Cfrac%7Br%5E4%7D%7B4%7D%5CBig%5Dd%5Ctheta)

![=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi](https://tex.z-dn.net/?f=%3D%5Cfrac%7B128%7D%7B3%7D%5CBig%5B%5Csin%5Ctheta-%5Ccos%5Ctheta%5CBig%5D_%7B0%7D%5E%7B2%5Cpi%7D%2B128%5Ctimes%202%5Cpi)
![=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi](https://tex.z-dn.net/?f=%3D%5Cfrac%7B128%7D%7B3%7D%5CBig%5B%5Csin%202%5Cpi-%5Ccos%202%5Cpi-%5Csin%200%2B%5Ccos%200%5CBig%5D%2B256%5Cpi)
Hence total charge is
coulombs per square meter.
Answer:
2
Step-by-step explanation:
So I'm going to use vieta's formula.
Let u and v the zeros of the given quadratic in ax^2+bx+c form.
By vieta's formula:
1) u+v=-b/a
2) uv=c/a
We are also given not by the formula but by this problem:
3) u+v=uv
If we plug 1) and 2) into 3) we get:
-b/a=c/a
Multiply both sides by a:
-b=c
Here we have:
a=3
b=-(3k-2)
c=-(k-6)
So we are solving
-b=c for k:
3k-2=-(k-6)
Distribute:
3k-2=-k+6
Add k on both sides:
4k-2=6
Add 2 on both side:
4k=8
Divide both sides by 4:
k=2
Let's check:
:


I'm going to solve
for x using the quadratic formula:







Let's see if uv=u+v holds.

Keep in mind you are multiplying conjugates:



Let's see what u+v is now:


We have confirmed uv=u+v for k=2.