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Help find the answer for problem number 4

Olin [163]

Adult ticket (a) = $5
Child ticket (c) = $2

785 tickets = $3280

a + c = 785 tickets
5a + 2c = $3280

c = 215 child tickets
a = 570 adult tickets

570 + 215 = 785 tickets
5(570) + 2(215) = $3280

There were 215 child tickets sold on Saturday

3 0

3 years ago

Slick Slack is a downloadable app that can be used to automatically generate excuses for being late to work based on the weather

Greeley [361]

Answer:

$(x_1,y_1) = (1,5840)$

$(x_2,y_2) = (2,13090)$

$y = 1812.5x + 4027.5$

$y_0 = 4027.5$ --- y intercept

Step-by-step explanation:

Given:

x = weeks; y = copies

The right parameters are:

After 1 week, 5840 copies were downloaded

After 5 weeks, 13090 copies were downloaded

So, the (x,y) forms are:

$(x_1,y_1) = (1,5840)$

$(x_2,y_2) = (5,13090)$

Next, we calculate the slope (m) using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{13090 - 5840}{5 - 1}$

$m = \frac{7250}{4}$

$m = 1812.5$

The equation is then calculated using:

$y = m(x - x_1) + y_1$

$y = 1812.5(x - 1) + 5840$

$y = 1812.5x - 1812.5 + 5840$

$y = 1812.5x + 4027.5$

To calculate the y intercept, substitute 0 for x

$y_0 = 1812.5*0 + 4027.5$

$y_0 = 0 + 4027.5$

$y_0 = 4027.5$

7 0

2 years ago

Robert works in a toll booth. About 18 to 23 cars pass through Robert's booth every 23 minutes. Which is a reasonable estimate f

BARSIC [14]

A reasonable number between 18 and 23 is 20. 69 minutes is 3 times 23 minutes, so we multiply the 20 by 3, getting an estimate of 60 cars.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

Katen [24]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

2 years ago

An airplane is traveling at 1,500 miles/minute and then 5 minutes later reaches a speed of 1800 miles/second. What is the averag

adell [148]

The answer to the problem is 60

8 0

3 years ago