Construct a line from A perpendicular to the y-axis, determine the distance from A to the y-axis along this perpendicular line, find a new point on the other side of the y-axis that is equidistant from the y-axis.
I believe this is the answer
Answer:
x=-2, y=-3
Step-by-step explanation:
we can rearrange the questions to make it easier
y=3x+3
2y=-3x-12
if you add the two equations, you get
3y=0x-9
so y=-9/3 = -3
plug y into one of the equations
-3=3x+3
-6=3x
x=-2
and then check your work by plugging into the other equation:
2*-3=-3(-2)-12
-6=6-12
-6=-6
Answer:
Ezra can water at most approximately 8 sunflower plants with the remaining amount of water.
Step-by-step explanation:
Given the inequality function
0.7S+0.5L≤11 where S represent the number of sunflower plants and L represent the number of lily plants Ezra's water supply can water, if Ezra waters 10 lily plants, then we can calculate the maximum amount of sunflower plant that he can water with the remaining amount of water by simply substituting L = 10 into the inequality function as shown;
0.7S+0.5L≤11
0.7S+0.5(10)≤11
0.7S+5≤11
Taking 5 to the other side:
0.7S≤11-5
0.7S≤6
S≤6/0.7
S≤8.57
This shows that Ezra can water at most approximately 8 sunflower plants with the remaining amount of water.
Answer: 
; 5
Step-by-step explanation:
Given series : ![[\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]](https://tex.z-dn.net/?f=%5B%5Cdfrac%7B5%7D%7B1%5Ccdot2%7D%5D%2B%5B%5Cdfrac%7B5%7D%7B2%5Ccdot3%7D%5D%2B%5B%5Cdfrac%7B5%7D%7B3%5Ccdot4%7D%5D%2B....%2B%5B%5Cdfrac%7B5%7D%7Bn%5Ccdot%28n%2B1%29%7D%5D)
Sum of series = ![S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]](https://tex.z-dn.net/?f=S_n%3D%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5C%20%5B%5Cdfrac%7B5%7D%7Bn%5Ccdot%28n%2B1%29%7D%5D%3D5%5B%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5Cdfrac%7B1%7D%7Bn%5Ccdot%28n%2B1%29%7D%5D)
Consider 
⇒ ![S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]](https://tex.z-dn.net/?f=S_n%3D5%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5Cdfrac%7B1%7D%7Bn%5Ccdot%28n%2B1%29%7D%3D5%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5B%5Cdfrac%7B1%7D%7Bn%7D-%5Cdfrac%7B1%7D%7Bn%2B1%7D%5D)
Put values of n= 1,2,3,4,5,.....n
⇒ 
All terms get cancel but First and last terms left behind.
⇒
Formula for the nth partial sum of the series :
Also, 
Answer:
5
Step-by-step explanation:
