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3 years ago

What is the method to the problem

Mathematics

1 answer:

madreJ [45]3 years ago

7 0

I think it might be the 3rd one.

Someone correct me if I'm wrong please.

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What is the area of Vivian’s garage door?

sergey [27]

The area of Vivian’s garage door is 62.5 square feet

<h3>Area of composite shape.</h3>

The Vivian garage is made up of a trapezoid and a rectangle. The area of the door is expressed as;

Area = Area of rectangle + area of trapezoid

Area = (5 * 8) + 0.5(7 + 8)*3
Area= 40 +22.5
Area = 62.5 square feet

Hence the area of Vivian’s garage door is 62.5 square feet

Learn more on composite shapes here: brainly.com/question/8370446

#SPJ1

3 0

1 year ago

Units of Capacity

marysya [2.9K]

Answer:

$\large\boxed{\large\boxed{0.2deciliter}}$

Explanation:

Build your conversion factors using the conversion table:

$1 cup = 0.237 liters\implies 1=0.237liters/1cup$

From the prefixes used in the metric system, you know that 1 deciliter = 0.100 liters. Then:

$1 deciliter = 0.100 liters\implies 1=1deciliter/0.100liters$

Arrange the expression to simplify the units and convert from cups to deciliters:

$9cups\times 0.237liters/cup\times 1deciliter/0.100liters=0.2133deciliter$

Round to the nearest tenth: 0.2 deciliter.

4 0

3 years ago

Read 2 more answers

Who knows how to solve these? I need help!!

Elina [12.6K]

1= 0.25
2=1.12
3= -075

4 0

3 years ago

Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago

Stats help please! Can’t figure out how to solve

mariarad [96]

This is unreadable...

8 0

2 years ago

Read 2 more answers