Revenue > Cost
R(x) > C(x)
0.50x > 10+0.20x
0.50x-0.20x > 10
0.30x > 10
x > 10/0.30
x > 33.333 which is approximate
Round up to the nearest whole number to get x = 34. You need to sell at least 34 cups of lemonade to have the revenue R(x) be larger than the cost C(x)
Answer: 34
Answer:
The quantity is 12
Step-by-step explanation:
Here, we want to find the solution to the equation.
Let the quantity we are looking for be x;
According to the question;
The quantity plus a quarter of the quantity = 15
x + 1/4(x) = 15
x + x/4 = 15
Multiply through by 4
4x + (x/4)*4 = 15 * 4
4x + x = 60
5x = 60
x = 60/5
x = 12
let's say the point dividing JK is say point P, so the JK segment gets split into two pieces, JP and PK
![\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ J(-25,10)\qquad K(5,-20)\qquad \qquad \stackrel{\textit{ratio from J to K}}{7:3} \\\\\\ \cfrac{J~~\begin{matrix} P \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} P \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~K} = \cfrac{7}{3}\implies \cfrac{J}{K} = \cfrac{7}{3}\implies3J=7K\implies 3(-25,10)=7(5,-20)\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Binternal%20division%20of%20a%20line%20segment%7D%20%5C%5C%5C%5C%5C%5C%20J%28-25%2C10%29%5Cqquad%20K%285%2C-20%29%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bratio%20from%20J%20to%20K%7D%7D%7B7%3A3%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7BJ~~%5Cbegin%7Bmatrix%7D%20P%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7B~~%5Cbegin%7Bmatrix%7D%20P%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~K%7D%20%3D%20%5Ccfrac%7B7%7D%7B3%7D%5Cimplies%20%5Ccfrac%7BJ%7D%7BK%7D%20%3D%20%5Ccfrac%7B7%7D%7B3%7D%5Cimplies3J%3D7K%5Cimplies%203%28-25%2C10%29%3D7%285%2C-20%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf P=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill\\\\ P=\left(\cfrac{(3\cdot -25)+(7\cdot 5)}{7+3}\quad ,\quad \stackrel{\textit{y-coordinate}}{\cfrac{(3\cdot 10)+(7\cdot -20)}{7+3}}\right) \\\\\\ P=\left( \qquad ,\quad \cfrac{30-140}{10} \right)\implies P=\left(\qquad ,~~\cfrac{-110}{10} \right)\implies P=(\qquad ,\quad -11)](https://tex.z-dn.net/?f=%5Cbf%20P%3D%5Cleft%28%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22x%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cquad%20%2C%5Cquad%20%5Cfrac%7B%5Ctextit%7Bsum%20of%20%22y%22%20values%7D%7D%7B%5Ctextit%7Bsum%20of%20ratios%7D%7D%5Cright%29%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20P%3D%5Cleft%28%5Ccfrac%7B%283%5Ccdot%20-25%29%2B%287%5Ccdot%205%29%7D%7B7%2B3%7D%5Cquad%20%2C%5Cquad%20%5Cstackrel%7B%5Ctextit%7By-coordinate%7D%7D%7B%5Ccfrac%7B%283%5Ccdot%2010%29%2B%287%5Ccdot%20-20%29%7D%7B7%2B3%7D%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20P%3D%5Cleft%28%20%5Cqquad%20%2C%5Cquad%20%5Ccfrac%7B30-140%7D%7B10%7D%20%5Cright%29%5Cimplies%20P%3D%5Cleft%28%5Cqquad%20%2C~~%5Ccfrac%7B-110%7D%7B10%7D%20%5Cright%29%5Cimplies%20P%3D%28%5Cqquad%20%2C%5Cquad%20-11%29)
<u>Answer:</u>
There are two approaches to solving the equation -3x + 10 = 4x - 20. The value of x after solving in both methods is 4.2857.
<u>Solution:</u>
Given, equation is -3x + 10 = 4x – 20.
We have to solve the given equation in two methods, now let us see the first method.
<u><em>1st method ⇒ by subtracting 4x from both sides.</em></u>
Then, -3x + 10 = 4x – 20 ⇒ -3x + 10 – 4x = 4x – 20 – 4x ⇒ -3x – 4x = -20 – 10 ⇒ -7x = -30 ⇒ 7x = 30
Then, x = 
Now, let us use the 2nd method.
<em><u>2nd method ⇒ by adding 3x on both sides.</u></em>
Then, -3x + 10 = 4x – 20 ⇒ -3x + 10 + 3x = 4x – 20 + 3x
⇒ 10 + 20 = 4x + 3x
⇒ 7x = 30
⇒ x = 
Hence, the value of x after solving in both methods is
= 4.2857.