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2 months ago

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A new outlet is being planned with an allocated budget of $50,000 per month for advertising and planned average price of $10.00

per pizza. Estimate the monthly sales (number of pizzas) of this outlet.

1 answer:

Ad libitum [116K]2 months ago

5 0

Answer:

5,000

$50000 \div 10 = 5000$

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The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

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<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

<em><u>Solution:</u></em>

Given that,

<em><u>The height of a ball thrown into the air after t seconds have elapsed is:</u></em>

$h = -16t^2 + 40t + 6$

<em><u>What is the first time, t, when the ball will reach a height of 20 feet?</u></em>

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

<em><u>Solve by quadractic formula</u></em>

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

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