Hi student, let me help you out! :)
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We are asked to find the two integers, given that they are consecutive, and their sum is 65.
- Consecutive integers are right next to each other, like 12 and 13. or 65 and 66.
Let the first integer be x, and let the second integer be x+1.
Their sum is 65. Let's set up our equation:
Combine like terms:
Subtract 1 from both sides of the equal sign:
Divide both sides by 2:
To find the second integer, subtract the first integer from the sum of the two integers:
The integers are: 33 and 32.
Hope it helps you out! :D
Ask in comments if any queries arise.
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Answer:
-2;12
Step-by-step explanation:
it's too easy!
Answer:
100$
Step-by-step explanation:
he profit from 1 popcorn=0.85$
from 200=170$
170$-70$=100
Check the picture below.
based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).
and that will give us two x-intercepts, at x = 3 and x = k.
since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.
now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.
let's find the y-intercept by setting x = 0 now,
![\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)](https://tex.z-dn.net/?f=%5Cbf%20y%3D0.5%28x-3%29%28x%2Bk%29%5Cimplies%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28x-3%29%28x%2Bk%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsetting%20x%20%3D%200%7D%7D%7By%3D%5Ccfrac%7B1%7D%7B2%7D%280-3%29%280%2Bk%29%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B1%7D%7B2%7D%28-3%29%28k%29%5Cimplies%20%5Cboxed%7By%3D-%5Ccfrac%7B3k%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20a%20triangle%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7Dbh%7D~~%20%5Cbegin%7Bcases%7D%20b%3D3%2Bk%5C%5C%20h%3Dy%5C%5C%20%5Cquad%20-%5Cfrac%7B3k%7D%7B2%7D%5C%5C%20A%3D1.5%5C%5C%20%5Cqquad%20%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7B3%7D%7B2%7D%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29)
now, we can plug those values on A = (1/2)bh,
![\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-2%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-2%29%5Cleft%28-%5Ccfrac%7B3%28-2%29%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%281%29%283%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20k%20%3D%20-1%7D%7D%7BA%3D%5Ccfrac%7B1%7D%7B2%7D%283%2Bk%29%5Cleft%28-%5Ccfrac%7B3k%7D%7B2%7D%20%5Cright%29%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283-1%29%5Cleft%28-%5Ccfrac%7B3%28-1%29%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7D%282%29%5Cleft%28%20%5Ccfrac%7B3%7D%7B2%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20A%3D1.5)
6 one-dollars = $6 15 five-dollars = $75 9 ten-dollars = $90 Add them
all up to get $171, so that is correct. Add the number of one-dollar
bills and the number of ten-dollar bills together. 6 + 9 = 15, which is
the number of five-dollar bills, so that is correct as well. Add all the
numbers of bills together, 6 + 9 + 15 = 15 + 15 = 30.
