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2 years ago

Could i have some quick quick help?

Mathematics

2 answers:

77julia77 [94]2 years ago

3 0

Turn over into vertex form

y=-30t²+450t-790
y=-30(t²-15x+79/3)

Solving

y=-30[(t-15/2)²-359/12]

Open brackets

y=-30(t-15/2)²+1795/2

Accurating

y=-30(t-7.5)²+897..5

Compare to Vertex form y=a(x-h)²+k

Vertex

(h,k)=7.5,897.5

Max profit is $897.5

ticket price should be $7.5

Hitman42 [59]2 years ago

3 0

Answer:

$7.50

Step-by-step explanation:

Completing the square formula

$\begin{aligned}y & =ax^2+bx+c\\& =a\left(x^2+\dfrac{b}{a}x\right)+c\\\\& =a\left(x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2\right)+c-a\left(\dfrac{b}{2a}\right)^2\\\\& =a\left(x-\left(-\dfrac{b}{2a}\right)\right)^2+c-\dfrac{b^2}{4a}\end{aligned}$

$\begin{aligned}P & =-30t^2+450t-790\\& =-30\left(t^2+\dfrac{450}{-30}t\right)-790\\\\& =-30\left(t^2+\dfrac{450}{-30}t+\left(\dfrac{450}{2(-30)}\right)^2\right)-790-(-30)\left(\dfrac{450}{2(-30)}\right)^2\\\\& =-30\left(t-\left(-\dfrac{450}{2(-30)}\right)\right)^2-790-\dfrac{450^2}{4(-30)}\\\\& =-30(t-7.5)^2+897.5\end{aligned}$

Therefore, the vertex is (7.5, 897.5)

So the ticket price that maximizes daily profit is the x-value of the vertex: $7.50

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Bess [88]

Answer: A. 5,000 meters

Step-by-step explanation:

Starting with the smallest unit which is option C. There are 10 millimeters in one centimetre. Option c in centimeters is:

= 30,000 / 10

= 3,000 meters.

Option c is smaller than B and so is incorrect.

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3 years ago

Tim predicts he will get 98 points on the next exam. If Tim actually earned 62 points, what was Tim's percent error? Round your

cupoosta [38]

% change= (new # - original #) ÷ original # x 100

original #= 98
new #= 62

% change= (62-98)/98 x 100
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ANSWER:
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3 years ago

Read 2 more answers

CALC- limits please show your method

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

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
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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

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

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3 years ago

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lina2011 [118]

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3 years ago