Answer:
it is 1/4 fourth or 0.25 or 0.250
7+10x-2x-2=-3
—-Combine like terms...10x-2x=8x and 7-2=5
8x+5=-3
——subtract 5 on both sides.. the 5’s will cancel each other and -3-5=-8
8x=-8
——Divide 8 on both sides...the 8’s will cancel out and -8 divided by 8 = -1
Answer: x = -1
Given Fig.is a triangle
Perimeter is simply sum of the all sides of a triangle
1/3x + 7/x^2 + (x+2)/x
=[ x+ 21 + 3x(x+2)]/3x^2
= (x+ 21 + 3x^2 + 6x)/3x^2
= (3x^2 + 7x +21)/3x^2
Answer:
This is a 10 percent increase
Step-by-step explanation:
Last week he spent 10 hours
This week 11 hours
percent increase = (new - original )/original * 100 percent
= (11-10)/10 * 100 percent
= 1/10 * 100 percent
= 10%
This is a 10 percent increase
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!