HELP!!!!!!!!!!!!!!!!!!!!!!!!

Which triangle’s unknown side length measures 7 units?

kow [346]

A (A right triangle with a side length of 5 and hypotenuse with length \sqrt{74})

8 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

4 years ago

What are the three choices equivalent to 27^4/3

12345 [234]

the answers are B,D and F

8 0

3 years ago

One of your company's salesmen estimates he will drive 20,000 miles this year and each of the next two years. As the service man

grigory [225]

Therefore, the total costs for year 1 will be $2,240, for year 2 they will be $2,040, and for year 3 they will be $2,120.

<h3><u /></h3><h3><u>Costs</u></h3>

Given that one of your company's salesmen estimates he will drive 20,000 miles this year and each of the next two years, and as the service manager for the company's fleet of six-cylinder compact automobiles, you are asked to determine the total costs for each of the three years, to determine said cost, the following calculation must be made:

Year 1:
20000 x 0.112 = X
2,240 = X

Year 2:
20000 x 0.102 = X
2,040 = X

Year 3:
20000 x 0.106 = X
2,120 = X

Therefore, the total costs for year 1 will be $2,240, for year 2 they will be $2,040, and for year 3 they will be $2,120.

Learn more about costs in brainly.com/question/4557688

4 0

3 years ago

The school store opened on the first day of school with 45 notebooks and 15 pencils. Within two days it sold all of these items.

malfutka [58]

The number of pencils sold on the first day is 5 pencils.

The number of notebooks sold on the first day is 10 notebooks.

The number of pencils sold on the second day 10 pencils.

The number of notebooks sold on the second day 35 notebooks.

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

Given data,

The total number of notebooks in the store = 45 notebooks.
The total number of Pencils in the store = 15 pencils.
The number of days that all items were sold = 2 days.

Now, you have to calculate the no.of notebooks and no. of pencils sold each day.

<u>In the first day :</u>

The number of sale of notebooks and pencils are given by,

Twice as many notebooks were sold as pencils.

Let us take, the number of pencils sold on the first day = x

And, the number of notebooks sold on the first day = 2x (Twice as pencils).

<u>In the second day :</u>

The number of sale of notebooks and pencils are given by,

For every 7 notebooks sold, 2 pencils were sold.

The number of pencils sold on the second day = 2y

The number of notebooks sold on the second day = 7y

<u> The equation is framed for number of notebooks sold on each day :</u>

The number of notebooks sold ⇒ 2x + 7y = 45 -------(1)

<u> The equation is framed for number of pencils sold on each day :</u>

The number of pencils sold: x + 2y = 15 ----------(2)

Solving the equations by multiplying eq(2) by 2 and subtract it from eq(1),

2x + 7y = 45

-<u> (2x + 4y) = 30</u>

⇒ y = 15/3

⇒ y = 5

The value of y is 5.

Substitute y=5 in eq (2),

⇒ x + 2(5) = 15

⇒ x + 10 = 15

⇒ x = 15 - 10

⇒ x = 5

The value of x is 5.

First day sale,

The number of pencils sold on the first day = x ⇒ 5 pencils

The number of notebooks sold on the first day = 2x ⇒ 10 notebooks

Second day sale,

The number of pencils sold on the second day = 2y ⇒ 10 pencils

The number of notebooks sold on the second day = 7y ⇒ 35 notebooks.

8 0

3 years ago