![\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%204x%2B3y%3D-8%5C%5C%5C%5C%20-8x-6y%3D16%20%5Cend%7Bcases%7D~%5Chspace%7B10em%7D%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}](https://tex.z-dn.net/?f=4x%2B3y%3D-8%5Cimplies%203y%3D-4x-8%5Cimplies%20y%3D%5Ccfrac%7B-4x-8%7D%7B3%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-8x-6y%3D16%5Cimplies%20-6y%3D8x%2B16%5Cimplies%20y%3D%5Ccfrac%7B8x%2B16%7D%7B-6%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B8%7D%7B-6%7Dx%2B%5Ccfrac%7B16%7D%7B-6%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D)
one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.
since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.
Answer:
43.35 years
Step-by-step explanation:
From the above question, we are to find Time t for compound interest
The formula is given as :
t = ln(A/P) / n[ln(1 + r/n)]
A = $2500
P = Principal = $200
R = 6%
n = Compounding frequency = 1
First, convert R as a percent to r as a decimal
r = R/100
r = 6/100
r = 0.06 per year,
Then, solve the equation for t
t = ln(A/P) / n[ln(1 + r/n)]
t = ln(2,500.00/200.00) / ( 1 × [ln(1 + 0.06/1)] )
t = ln(2,500.00/200.00) / ( 1 × [ln(1 + 0.06)] )
t = 43.346 years
Approximately = 43.35 years
X=cost per hamburgurs
y=cost per hot dog
200x+150y=1450
200x+250y=1750
we can multiply first equation by -1 and add to 2nd
-200x-150y=-1450
<u>200x+250y=1750 +</u>
0x+100y=300
100y=300
divide both sides by 100
y=3
sub back
200x+150y=1450
200x+150(3)=1450
200x+450=1450
minus 450 both sides
200x=1000
divide both sides by 200
x=5
the cost of a hamburgur is $5
the cost of a hot dog is $3
Answer:
Step-by-step explanation:
just replace x by 2
Answer:
total population in the year 2000=4400
total population in the year 2010=5060
no of people increased=5060-4400
=660
increase percent=660/4400*100%
=3/20*100%
=15%
Step-by-step explanation:
15/100*5060
0.15*5060
759
in 2020 population will be
5060+759
5819
therefore the population will be 5819 in 2020
