Hello!
Looking for a Genius Answer? Here It Is!
To get the solution, we are looking for, we need to point out what we know.
1. We assume, that the number 150 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 100% equals 150, so we can write it down as 100%=150.
4. We know, that x% equals 25 of the output value, so we can write it down as x%=25.
5. Now we have two simple equations:
1) 100%=150
2) x%=25
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=150/25
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for 25 is what percent of 150
100%/x%=150/25
(100/x)*x=(150/25)*x - we multiply both sides of the equation by x
100=6*x - we divide both sides of the equation by (6) to get x
100/6=x
16.6666666667=x
x=16.6666666667
now we have:
<span>25 is 16.6666666667% of 150
</span>
Hope this Helps! Took Me Time! Have A Wonderful Day! :)
Answer: z = 3
Step-by-step explanation: To solve this equation for <em>z</em>, we can first combine our like terms on the left side of the equation. Since 12 and 7 both have <em>z</em> after their coefficient, we can subtract 12z - 7z to get 5z.
Now we have 5z - 2 = 13.
To solve from here, we add 2 to the left side of the equation in order to isolate 5z. If we add 2 to the left side, we must also add 2 to the right side. On the left side, the -2 and +2 cancel out. On the right, 13 + 2 simplifies to 15.
Now we have 5z = 15.
Solving from here, we divide both sides of the equation by 5 to get <em>z</em> alone. On the left side, the 5's cancel out and we are simply left with <em>z</em>. On the right side, 15 divided by 5 simplifies to 3 so we have z = 3.
Answer:
3pf² - 21p²f + 6pf - 42p²
= 3pf² + 6pf - 21p²f - 42p²
= 3pf(f +2) - 21p²(f + 2)
= (3pf - 21p²)(f +0 2)
Answer: (3pf - 21p²)(f + 2)
Step-by-step explanation:
Answer:1002
Step-by-step explanation:
and 
and 
as 
Applying this we get
![\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csum_%7B1%7D%5E%7B1002%7D%5Cleft%20%5B%20%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7Bk%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%2B%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7B%282005-k%29%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%5Cright%20%5D)
every 
there exist 
such that 
therefore
![\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csum_%7B1%7D%5E%7B1002%7D%5Cleft%20%5B%20%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7Bk%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%2B%5Ccos%5E2%5Cleft%20%28%20%5Cfrac%7B%282005-k%29%5Cpi%20%7D%7B2%5Ccdot%202005%7D%5Cright%20%29%5Cright%20%5D%3D1002)
17-6 x 10/2+ 12
17-60/2+12
17-30+12
17-18
= -1
