First solve for the slope, m using the two points given. It doesn't matter which point you choose as point 1 or 2 as long as you're consistent.
m = (y2 - y1)/(x2 - x1)
point 1: (–6.4, –2.6)
point 2: (5.2, 9)
m = (9 - -2.6)/(5.2 - -6.4)
m = (9 + 2.6)/(5.2 + 6.4)
m = 11.6/11.6
m = 1
put the newly found slope into the linear equation for m
y = (1)x + b
y = x + b
Now solve for the y-intercept, b
by putting one of the given points
9 = 5.2 + b
b = 9 - 5.2
b = 3.8
final equation:
y = x + 3.8
Answer:(-8,-4)
Step-by-step explanation:
To make it easier, you calculate the volume of the first aquarium.
1st aquarium:
V = L x W x H
V = 8 x 9 x 13
V = 72 x 13
V = 936 in.
Rate: 936 in./2 min.
Now that you've got the volume and rate of the first aquarium, you can find how many inches of the aquarium is filled within a minute, which is also known as the unit rate. To do that, you have to divide both the numerator and denominator by their least common multiple, which is 2. 936 divided by 2 is 468 and 2 divided by 2 is 1.
So the unit rate is 468 in./1 min. Now that you've got the unit rate, you can find out how long it'll take to fill the second aquarium up by finding its volume first.
2nd aquarium:
V = L x W x H
V = 21 x 29 x 30
V = 609 x 30
V 18,270 inches
Calculations:
Now, you divide 18,270 by 468 to find how many minutes it will take to fill up the second aquarium. 18,270 divided by 468 is about 39 (the answer wasn't exact, so I said "about").
2nd aquarium's rate:
18,270 in./39 min.
As a result, it'll take about 39 minutes to fill up an aquarium measuring 21 inches by 29 inches by 30 inches using the same hose. I really hope I helped and that you understood my explanation! :) If I didn't, I'm sorry. I tried. :(
Answer:
4. sin 40= 0.7451131604793488384
5. cos 40= -0.666938061652261888
6. tan 50= -0.2719006119976306688
7. tan 40= -1.1172149309238962176
8. sin A= -0.5882162892399388672
9. cos A= -0.8087036521945458688
10. tan A= 0.7273570332515760128
now, if we take 2000 to be the 100%, what is 2200? well, 2200 is just 100% + 10%, namely 110%, and if we change that percent format to a decimal, we simply divide it by 100, thus 
.
so, 1.1 is the decimal number we multiply a term to get the next term, namely 1.1 is the common ratio.
![\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\a_1=2000\\r=1.1\\n=4\end{cases}\\\\\\S_4=2000\left[ \cfrac{1-(1.1)^4}{1-1.1} \right]\implies S_4=2000\left(\cfrac{-0.4641}{-0.1} \right)\\\\\\S_4=2000(4.641)\implies S_4=9282](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bsum%20of%20a%20finite%20geometric%20sequence%7D%5C%5C%5C%5CS_n%3D%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7D%5C%20a_1%5Ccdot%20r%5E%7Bi-1%7D%5Cimplies%20S_n%3Da_1%5Cleft%28%20%5Ccfrac%7B1-r%5En%7D%7B1-r%7D%20%5Cright%29%5Cquad%20%5Cbegin%7Bcases%7Dn%3Dn%5E%7Bth%7D%5C%20term%5C%5Ca_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5Cr%3D%5Ctextit%7Bcommon%20ratio%7D%5C%5C----------%5C%5Ca_1%3D2000%5C%5Cr%3D1.1%5C%5Cn%3D4%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5CS_4%3D2000%5Cleft%5B%20%5Ccfrac%7B1-%281.1%29%5E4%7D%7B1-1.1%7D%20%5Cright%5D%5Cimplies%20S_4%3D2000%5Cleft%28%5Ccfrac%7B-0.4641%7D%7B-0.1%7D%20%20%5Cright%29%5C%5C%5C%5C%5C%5CS_4%3D2000%284.641%29%5Cimplies%20S_4%3D9282%20)
