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3 years ago

!! HELP ASAP !!

Mathematics

1 answer:

kherson [118]3 years ago

6 0

Can you provide a picture?

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A company that offers tubing trips down a river rents tubes for a person to use for $20 and “cooler” tubes to carry food and wat

aivan3 [116]

The group rent 11 river rent tubes and 4 cooler tubes.

Step-by-step explanation:

Rent for river rent tube = $20

Rent for cooler tubes = $12.50

Total spent = $270

Total tubes rented = 15

Let,

x be the number of river rent tube.

y be the number of cooler tubes.

According to given statement;

x+y=15 Eqn 1

20x+12.50y=270 Eqn 2

Multiplying Eqn 1 by 20

$20(x+y=15)\\20x+20y=300\ \ \ Eqn\ 3$

Subtracting Eqn 2 from Eqn 3

$(20x+20y)-(20x+12.50y)=300-270\\20x+20y-20x-12.50y=30\\7.50y=30$

Dividing both sides by 7.5

$\frac{7.5y}{7.5}=\frac{30}{7.5}\\y=4$

Putting y=4 in Eqn 1

$x+4=15\\x=15-4\\x=11$

The group rent 11 river rent tubes and 4 cooler tubes.

Keywords: linear equations, subtraction

Learn more about linear equations at:

#LearnwithBrainly

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3 years ago

You purchase a painting for 9,000 that has an appreciation rate of 12%. What is the growth

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Answer:

Step-by-step explanation:

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4. Find the surface area of the space figure represented by the net.

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The correct answer is C. 20 characters

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Simplify completely 7x^2+140x+700/x^2+10x-24

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Answer: 707x^2+150x-24

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3 years ago

The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas

saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by $s(t) = -t^3+3t^2+7t +4$ , where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

$s(t) = -t^3+3t^2+7t +4$

Derivate w.r.t. time,

$v(t)=s'(t) = -3t^2+6t+7$

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

$x=-\frac{b}{2a}$

Where, a=-3, b=6 and c=7

$t=-\frac{6}{2(-3)}$

$t=-\frac{6}{-6}$

$t=1$

As 1 lie between interval [0,4]

Substitute t=1 in the function,

$v(t)= -3(1)^2+6(1)+7$

$v(t)= -3+6+7$

$v(1)=10$

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

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3 years ago

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