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3 years ago

The unit rate of 2 for $3.33

Mathematics

1 answer:

frozen [14]3 years ago

8 0

Step-by-step explanation:

3.33 divided by 2 for unit rate of $1.655

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A square prism measuring 4 ft along each edge of the base and 6ft tall

ki77a [65]

4ft * 4ft * 6ft = 96ft squared

6 0

2 years ago

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago

a piece of land 120 ft by 240 ft contains a circular garden. surrounding the garden, there is a circular fence that is 20 ft in

damaskus [11]

the area of the piece of land that is not part of the garden is $28,486ft^2$ .

<u>Step-by-step explanation:</u>

Here we have , a piece of land 120 ft by 240 ft contains a circular garden. surrounding the garden, there is a circular fence that is 20 ft in diameter. We need to find what is the area of the piece of land that is not part of the garden . Let's find out:

Let's calculate area of land and circular garden , and in order to find the area of the piece of land that is not part of the garden we will subtract area of circular garden from land !

Area of Land:

Area of land is given by

⇒ $length(width)$

⇒ $120(240)$

⇒ $28,880ft^2$

Area of Garden:

Area of circular garden given by

⇒ $\pi r^2$

⇒ $\pi (\frac{D}{2})^2$

⇒ $\pi (\frac{20}{2})^2$

⇒ $\pi (10)^2$

⇒ $100\pi$

⇒ $100(3.14)$

⇒ $314ft^2$

Hence , the area of the piece of land that is not part of the garden is 28,880-314 = 28,486 . Therefore , the area of the piece of land that is not part of the garden is $28,486ft^2$ .

8 0

2 years ago

Put the following products in order from greatest to least 6x 2/5 , 3 2/3 x 2/5, 2/7 X 2/5, 2/2 x2/5

DedPeter [7]

6x2/5>3 2/3x 2/5>2/2x2/5>2/7x2/5

6 0

3 years ago

A yearbook page will show 20 photos displayed in rows.Each row must contain the same number of photos.How many different ways ca

gregori [183]

Answer:

There are five different ways, in which we can arrange the photos.

Step-by-step explanation:

There are five different ways to arrange the photos.

1. 5 photos in each 4 rows.

2. 4 photos in each 5 rows.

3. 2 photos in each 10 rows.

4. 10 photos in each 2 rows.

5. 20 photos in a single row.

5 0

3 years ago

The unit rate of 2 for $3.33 ​

The unit rate of 2 for $3.33