The statement best describes Cheryl's computer is option C. Cheryl accelerated to 65 mph, drove at a constant speed for 5.5 minutes, and then decelerated to 45 mph.
<h3>How to find the function which was used to make graph?</h3>
A graph contains data of which input maps to which output.
Analysis of this leads to the relations which were used to make it.
If we know that the function crosses the x-axis at some point, then for some polynomial functions, we have those as roots of the polynomial.
Let's assume the graph of Cheryl's commute was like the one below.
We can see that she started at 0 mph.
One minute later, she was up to 65 mph, so she had accelerated (increased her speed).
At 6.5 min (5.5 min later) her speed was still 65 mph therefore, she was driving at a constant speed.
Over the next 2.5 min, her speed dropped to 45 mph, therefore she was decelerating.
Learn more about finding the graphed function here:
brainly.com/question/27330212
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Answer:
amigo mío la respuesta es la letra B
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)
In order from least to greatest:
5/10 , 7/12 , 4/6
You have to find a common denominator of all 3 numbers which is 60 and then you proceed with the next steps.
Writing an equation is the easiest way to figure this out.
b = boys, g = girlsb = 3/4g35 students = g + b35 = g + 3/4g35 = 7/4g35/7/4 = g20 girls35 - 20 = b or 3/4(35) = b15 boys
