Answer:
pencils amount = 9 pens amount= 9
Step-by-step explanation:
We start this problem forming two linear equations system.
3*X+7*Y=90
Where X is the amount of pencils and Y is the amount of pens
The second equation is given by the statement, the amount of pencils is equal to the amount of pens, so the second equation is:
X=Y
Merging both equations we have:
3*X+7*X=90
Solving for X we have:
10*X=90
X=9
X=Y=9
So,
Finding problems such as "x is 10% of y" and solving them is really easy: simply multiply the number you are given by 10 to get 100% of that number.
3.3 * 10 = 33
The number you are looking for is 33.
So lets say u have 100$ and it decreases by 100% u would have no money left
Answer:
the answer is No
Step-by-step explanation:
Since the first line has a slope of -4/7 and the other line has a slope of -1/7. This can be shown in point slope form y-y1=m(x-x1) where m=∆y/∆x.
Through points P and Q, the point slope is represented as -1-(-5)=m(-5-2) ; 4 = -7m ; m = 4/-7. The next two points R and S can be shown in point slope as 5-4=m(-2-5) ; 1 = -7m ; m =1/-7. With the fact that parallel lines must have equal slopes, these two lines cannot be parallel because -4/7≠-1/7
Given two point charges
![q_1](https://tex.z-dn.net/?f=q_1)
and
![q_2](https://tex.z-dn.net/?f=q_2)
which are a distance
![r_{12}](https://tex.z-dn.net/?f=r_%7B12%7D)
apart.
The force
![\overrightarrow{F}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D)
on one of the charges is proportional to the magnitude of its own charge and inversely proportional to the square of the distance between them.
i.e.
![\overrightarrow{F}= \frac{k_e|q_1q_2|}{r_{12}^2}](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%3D%20%5Cfrac%7Bk_e%7Cq_1q_2%7C%7D%7Br_%7B12%7D%5E2%7D%20)
where
![k_e](https://tex.z-dn.net/?f=k_e)
is the constant of proportionality
![8.99\times10^9 \ Nm^2C^{-2}](https://tex.z-dn.net/?f=8.99%5Ctimes10%5E9%20%5C%20Nm%5E2C%5E%7B-2%7D)
Given that
![q_1=2.0\mu C](https://tex.z-dn.net/?f=%3Cspan%3Eq_1%3D2.0%5Cmu%20C)
and
![q_2=4.0\mu C](https://tex.z-dn.net/?f=q_2%3D4.0%5Cmu%20C)
Also, given that
![r_ 1 = (4.0i-2.0 j+5.0k)](https://tex.z-dn.net/?f=r_%201%20%3D%20%284.0i-2.0%20j%2B5.0k%29)
m and
![r_2 = (8.0 i+ 5.0 j-9.0)](https://tex.z-dn.net/?f=r_2%20%3D%20%288.0%20i%2B%205.0%20j-9.0%29)
m, then
![|r_{12}|=|r_2-r_1| \\ \\ =|(8.0 i+ 5.0 j-9.0)-(4.0i-2.0 j+5.0k)| \\ \\ =|4.0i+7.0j-14k|= \sqrt{4^2+7^2+(-14)^2} = \sqrt{16+49+196} \\ \\ = \sqrt{261} =16.2 \ m](https://tex.z-dn.net/?f=%7Cr_%7B12%7D%7C%3D%7Cr_2-r_1%7C%20%5C%5C%20%20%5C%5C%20%3D%7C%288.0%20i%2B%205.0%20j-9.0%29-%284.0i-2.0%20j%2B5.0k%29%7C%20%5C%5C%20%20%5C%5C%20%3D%7C4.0i%2B7.0j-14k%7C%3D%20%5Csqrt%7B4%5E2%2B7%5E2%2B%28-14%29%5E2%7D%20%3D%20%5Csqrt%7B16%2B49%2B196%7D%20%20%5C%5C%20%20%5C%5C%20%3D%20%5Csqrt%7B261%7D%20%3D16.2%20%5C%20m)
Therefore, <span>the force of
![q_2](https://tex.z-dn.net/?f=q_2)
on
![q_1](https://tex.z-dn.net/?f=q_1)
is given by
![\overrightarrow{F}= \frac{8.99\times10^9\cdot2\times10^{-6}\cdot4\times10^{-6}}{(16.2)^2} \\ \\ = \frac{7.192\times10^{-5}}{261} =2.76\times10^{-7} \ N](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%3D%20%5Cfrac%7B8.99%5Ctimes10%5E9%5Ccdot2%5Ctimes10%5E%7B-6%7D%5Ccdot4%5Ctimes10%5E%7B-6%7D%7D%7B%2816.2%29%5E2%7D%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B7.192%5Ctimes10%5E%7B-5%7D%7D%7B261%7D%20%3D2.76%5Ctimes10%5E%7B-7%7D%20%5C%20N)
</span>