Debbie sells cookies and hotdogs. The daily cost of making cookies is $460 more than the difference between the square of the number of cookies sold and 20 times the number of cookies sold. The daily cost of making hotdogs is modeled by the following equation: C(x) = 2x2 - 80x + 1,200 C(x) is the cost in dollars of selling x hotdogs. Which statement best compares the minimum daily cost of making cookies and hotdogs? It is greater for hotdogs than cookies because the approximate minimum cost is $300 for cookies and $392 for hotdogs. It is greater for hotdogs than cookies because
Here are the answer choices: It is greater for hotdogs than cookies because the approximate minimum cost is $300 for cookies and $392 for hotdogs. It is greater for hotdogs than cookies because the approximate minimum cost is $360 for cookies and $400 for hotdogs. It is greater for cookies than hotdogs because the approximate minimum cost is $400 for cookies and $360 for hotdogs. It is greater for cookies than hotdogs because the approximate minimum cost is $392 for cookies and $300 for hotdogs.
@jdoe0001
@sjada143
well... what do you think about the 1st equation? any ideas what it might look like? the cost of making cookies that is
If I had to go with one answer it would be "B"
in short.. you have 2 equations... ... so you'd want to find their vertex... then you can tell which vertex is higher and from what ranges
Yea bro that's the problem I dont know how to make the equation
difference => minus - so "$460 more than " + 460 "the difference between " ( ? - ? ) + 460 "difference between the square of the number of cookies " \((x^2 - ?)\) "and 20 times the number of cookies" 20 * x = > 20x "the difference between the square of the number of cookies sold and 20 times the number of cookies" \(\bf (x^2 - 20x))\) ----------------------------------------------------- "$460 more than the difference between the square of the number of cookies sold and 20 times the number of cookies" \(\bf x^2-20x+460\)
Oh
Ok now what do I do
so you have 2 quadratic equtions, and thus 2 parabolas.. \(\large { \textit{vertex of a parabola}\\ \quad \\ \begin{array}{llll} f(x) = {\color{red}{ 1}}s^2{\color{blue}{ -20}}s{\color{green}{ +460}}\\ c(x) = {\color{red}{ 2}}s^2{\color{blue}{ -80}}s{\color{green}{ +1,200}} \end{array}\ \quad \left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right) }\) find their vertex... and then do a quick graph once you know their vertex.... you can see where the price goes up and down and where it makes a U-turn
hmmm should be "x" there anyhow \(\large { \textit{vertex of a parabola}\\ \quad \\ \begin{array}{llll} f(x) = {\color{red}{ 1}}x^2{\color{blue}{ -20}}x{\color{green}{ +460}}\\ c(x) = {\color{red}{ 2}}x^2{\color{blue}{ -80}}x{\color{green}{ +1,200}} \end{array}\ \quad \left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right) }\)
Ok, I found out that the vertex of the first equation is 20,400 and the second equation's vertex is 10,360
well... ok... so now we know where the costs DROPS and then goes back UP
so... what do you think? keep in mind that the vertex....is an upward opening parabola, since the leading term is positive the vertex is where the cost of making either cookies or hotdogs drops to its lowest and if you're making them... of course you'd want the cost low
Ok so it is apparenly B when I graph it
@jdoe0001 ?
@jdoe0001 Are you there?
yes
@JuliusTheGreat @StudyGurl14
Anyone?
Oh thank god your back:D
@jdoe0001 I found out it is B, am I right?
you're correct the cost of hotdogs at its minimum is 400 the cost of cookies at its minimum is 360 thus the cost for hotdogs is greater
sit is a bit lagged, also
Ok can you help with one more?
site that is, is a bit lagged =)
sure, post anew, if I dunno somene else may :)
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