We have,
x+y:x-y=3:2
To prove that, x+ y : x - y = 3:2.
x+y=3
And,
x+y = 2
Adding (1) and (2), we get
x+y+x-y=3+2
⇒2x=5
⇒ x =
Put x = in (1), we get
+y=3
⇒ y=3-=
x+y:x-y=
=3:2
So, x+y:x-y=3:2.
Answer:
+ 2
Step-by-step explanation:
Group:
(2
+ 4x) (-5
- 10)
2x (
+ 2) -5 (
+ 2)
Common binomial factor = (
+ 2)
Answer:
0 with multiplicity 3, 3 with multiplicity 1, and 6 with multiplicity 1.
Answer:
![R=\frac{QJ}{I^2t}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BQJ%7D%7BI%5E2t%7D)
Step-by-step explanation:
So we have the equation:
![Q=\frac{I^2Rt}{J}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7BI%5E2Rt%7D%7BJ%7D)
And we want to solve for R.
First, let's multiply both sides by J to remove the fraction on the right. So:
![(J)Q=(J)\frac{I^2Rt}{J}](https://tex.z-dn.net/?f=%28J%29Q%3D%28J%29%5Cfrac%7BI%5E2Rt%7D%7BJ%7D)
Simplify the right:
![JQ=I^2Rt](https://tex.z-dn.net/?f=JQ%3DI%5E2Rt)
We can rewrite our equation as:
![JQ=R(I^2t)](https://tex.z-dn.net/?f=JQ%3DR%28I%5E2t%29)
So, to isolate the R variable, divide both sides by I²t:
![\frac{JQ}{I^2t}=\frac{R(I^2t)}{I^2t}](https://tex.z-dn.net/?f=%5Cfrac%7BJQ%7D%7BI%5E2t%7D%3D%5Cfrac%7BR%28I%5E2t%29%7D%7BI%5E2t%7D)
The right side cancels, so:
![R=\frac{QJ}{I^2t}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BQJ%7D%7BI%5E2t%7D)
And we are done!