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GarryVolchara [31]

2 years ago

7

Maria wants to store her soup in a cylindrical container and completely cover it in aluminum foil. What is the surface area of t

his container? Use 3 for π.

1 answer:

ioda2 years ago

6 0

Answer:

I need the measurements to calculate the surface area

Step-by-step explanation:

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Write the equation that shows 40 is 8 times more than 5

Vlad1618 [11]

Answer:

8X5=40

Step-by-step explanation:

3 0

2 years ago

Find SV. It’s hard for me :(

rosijanka [135]

Answer:

56

Step-by-step explanation:

from theorem , VR/SR=VU/UT

5 0

2 years ago

Solve for c.<br><br> 3 − 7c − 20c = 7(–7c − 19) + 14c<br><br> c =

OleMash [197]

Answer:

$\boxed{c = -17}$

Step-by-step explanation:

$= > 3 - 7c - 20c = 7( - 7c - 19) + 14c \\ \\ = > 3 - 27c = ( - 7c \times 7) - (7 \times 19) + 14c \\ \\ = > 3 - 27c = - 49c - 133 + 14c \\ \\ = > 3 - 27c = - 35c - 133 \\ \\ = > 3 - 27c + 35c = - 133 \\ \\ = > 3 + 8c = - 133 \\ \\ = > 8c = - 133 - 3 \\ \\ = > 8c = - 136 \\ \\ = > c = - \frac{136}{8} \\ \\ = > c = - 17$

7 0

3 years ago

Read 2 more answers

Please help have only an hour please use picture

____ [38]

Answer:

The answer is C. n = 3

Step-by-step explanation:

I got this answer by using the butterfly method. I multiplied 3 and 2 and got 6. Then take 6 and divide it by 2 and you get 3.

So therefore, n = 3

Please mark brainliest if this helped.

4 0

2 years ago

Read 2 more answers

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