To set up your system of equations is to have 2 equations. They both would have 1 similar variable. For example, Jenny is 5 more her brother's age. She is also 2 more than twice her brother's age. You would set this up as Y, which is Jenny, and X, her brother. Y = 5 + x and Y = 2x + 2. Make sure you put Y in the same spot in both equations. Then take the other half of the equations and simplify them. 5 + x = 2x + 2. Subtract x from both sides to get 5 = x + 2. Subtract 2 from both sides to get 3 = x. Now replace the X from both equations to get Y = 5+3, and Y = 2(3) + 2. You will see both add up to 8, which is how old Jenny is in this example. Hope this helped!
Check the picture below.
let's firstly convert the mixed fractions to improper fractions.
![\stackrel{mixed}{7\frac{1}{2}}\implies \cfrac{7\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{15}{2}} ~\hfill \stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B7%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B7%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B15%7D%7B2%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B12%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B12%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B25%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\stackrel{\textit{\Large Areas}}{\stackrel{two~triangles}{2\left[ \cfrac{1}{2}\left(\cfrac{15}{2} \right)(10) \right]}~~ + ~~\stackrel{\textit{three rectangles}}{(10)(15)~~ + ~~\left( \cfrac{15}{2} \right)(15)~~ + ~~\left( \cfrac{25}{2} \right)(15)}} \\\\\\ 75~~ + ~~150~~ + ~~112.5~~ + ~~187.5\implies \boxed{525}](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btwo~triangles%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%5Cleft%28%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2810%29%20%5Cright%5D%7D~~%20%2B%20~~%5Cstackrel%7B%5Ctextit%7Bthree%20rectangles%7D%7D%7B%2810%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B15%7D%7B2%7D%20%5Cright%29%2815%29~~%20%2B%20~~%5Cleft%28%20%5Ccfrac%7B25%7D%7B2%7D%20%5Cright%29%2815%29%7D%7D%20%5C%5C%5C%5C%5C%5C%2075~~%20%2B%20~~150~~%20%2B%20~~112.5~~%20%2B%20~~187.5%5Cimplies%20%5Cboxed%7B525%7D)
Total slices=4*6=24. Take away 3 slices leaving 21=18+3=3 pizzas and ½ a pizza, 3½.
Answer:
15.6
Step-by-step explanation:
First, multiply the midpoint of each class by its frequency, as follows:
Class midpoint frequency midpoint*frequency
0-9 4.5 24 4.5*24 = 108
10-19 14.5 20 14.5*20 = 290
20-29 24.5 32 24.5*32 = 784
Total 76 1182
The mean is computed as the division between the addition of the "midpoint*frequency" column by the addition of "frequency" column.
mean = 1182/76 ≈ 15.6
Answer:
The answer is B: (1,2)
Step-by-step explanation:
(solved using elimination method)
-2x+y=0
y=x+1
-x+y=1
-2x+y=0
x+y=-1
-x=-1
x=1
-1+y=1
y=2
(x,y)=(1,2)
