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)
Answer:
y = 2/3x + 19/3
Step-by-step explanation:
First, this is your current equation: y = 2/3x + b. Plug in your point to the equation, it should look like this: 5 = 2/3(-2) + b. 2/3 times -2 equals -4/3. So, this is what your equation should look like: 5 = -4/3 + b. Add 4/3 to both sides of the equation to get 19/3 = b. Go back to your original equation and plug 19/3 to b. This is your final equation: y = 2/3x + 19/3. Hope this helped!
Answer:
sorry if image isn't clear, tried many shots
Step-by-step explanation:
Answer:
ok so if equilateral triangle has 1 inch and box has 9 inch of height, the lateral area would be 27 square-inch
Step-by-step explanation:
