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arlik [135]
3 years ago
7

If someone could help me out with this, I would greatly appreciate it!!

Mathematics
1 answer:
kati45 [8]3 years ago
7 0

We are given that revenue of Tacos is given by the mathematical expression -7x^{2}+32x+240.

(A) The constant term in this revenue function is 240 and it represents the revenue when price per Taco is $4. That is, 240 dollars is the revenue without making any incremental increase in the price.

(B) Let us factor the given revenue expression.

-7x^{2}+32x+240=-7x^{2}+60x-28x+240\\-7x^{2}+32x+240=x(-7x+60)+4(-7x+60)\\-7x^{2}+32x+240=(-7x+60)(x+4)\\

Therefore, correct option for part (B) is the third option.

(C) The factor (-7x+60) represents the number of Tacos sold per day after increasing the price x times. Factor (4+x) represents the new price after making x increments of 1 dollar.

(D) Writing the polynomial in factored form gives us the expression for new price as well as the expression for number of Tacos sold per day after making x increments of 1 dollar to the price.

(E) The table is attached.

Since revenue is maximum when price is 6 dollars. Therefore, optimal price is 6 dollars.


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gladu [14]
A. Factor the numerator as a difference of squares:

\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6

c. As x\to\infty, the contribution of the terms of degree less than 2 becomes negligible, which means we can write

\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4

e. Let's first rewrite the root terms with rational exponents:

\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}

Next we rationalize the numerator and denominator. We do so by recalling

(a-b)(a+b)=a^2-b^2
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In particular,

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(x^{1/2}-x)(x^{1/2}+x)=x-x^2

so we have

\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}

For x\neq0 and x\neq1, we can simplify the first term:

\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x

So our limit becomes

\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43
3 0
3 years ago
8/25 x 7/8 Please help guys I am almost done with this paper and I am tired
Oduvanchick [21]
Hopes this helps:

Answer: 7/25

1. Cancel 8.

1/25 • 7

2. Simplify.

7/25
4 0
3 years ago
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