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3 years ago

Admission prices at a football game were $5 for adults and $ 2 for children.

Mathematics

1 answer:

satela [25.4K]3 years ago

8 0

Answer:

404 adult tickets were sold, and 250 child tickets.

Step-by-step explanation:

This question can be solved by a system of equations.

I am going to say that

x is the number of adult tickets sold.

y is the number of child tickets sold.

654 tickets were sold.

This means that

$x + y = 654$

$y = 654 - x$

Admission prices at a football game were $5 for adults and $ 2 for children. The total value of the tickets sold was $2520.

This means that

$5x + 2y = 2520$

Since $y = 654 - x$

$5x + 2(654 - x) = 2520$

$5x + 1308 - 2x = 2520$

$3x = 1212$

$x = \frac{1212}{3}$

$x = 404$

$y = 654 - x = 654 - 404 = 250$

404 adult tickets were sold, and 250 child tickets.

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NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.

Oliga [24]

Answer:

$\displaystyle 1)48.2 \: \: \text{sec}$

$\rm \displaystyle 2)3021.6 \: m$

Step-by-step explanation:

<h3>Question-1:</h3>

so when flash down occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,

to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:

$\displaystyle - 4.9 {t}^{2} + 229t + 346 = 0$

to solve the equation can consider the quadratic formula given by

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

so let our a,b and c be -4.9,229 and 346 Thus substitute:

$\rm\displaystyle t = \frac{ - (229) \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

remove parentheses:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

simplify square:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)}$

simplify multiplication:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 6781.6} }{ - 9.8}$

simplify Substraction:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 45659.4} }{ - 9.8}$

by simplifying we acquire:

$\displaystyle t = 48.2 \: \: \: \text{and} \quad - 1.5$

since time can't be negative

$\displaystyle t = 48.2$

hence,

at 48.2 seconds splashdown occurs

<h3>Question-2:</h3>

to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered

$\displaystyle x _{ \text{max}} = \frac{ - b}{2a}$

let a and b be -4.9 and 229 respectively thus substitute:

$\displaystyle t _{ \text{max}} = \frac{ - 229}{2( - 4.9)}$

simplify which yields:

$\displaystyle t _{ \text{max}} = 23.4$

now plug in the maximum t to the function:

$\rm \displaystyle h(23.4)- 4.9 {(23.4)}^{2} + 229(23.4)+ 346$

simplify:

$\rm \displaystyle h(23.4) = 3021.6$

hence,

about 3021.6 meters high above sea-level the rocket gets at its peak?

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Answer:

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sin(E) = opposite-leg / hypotenuse = 3 / 5

Answer: 3 / 5

Step-by-step explanation:

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