All categories

3 years ago

HELP ME IM GONNA LOSE MY MIND IN MINUTES

Mathematics

2 answers:

VLD [36.1K]3 years ago

7 0

Oh gosh that’s hard try using a calculator if you can

Nonamiya [84]3 years ago

5 0

Answer:

I already lost my mind use a calculator that all I got for you sorry

You might be interested in

A company leases a variety of office and production floor machines for an annual cost of $70,476. What are the leasing costs per

DochEvi [55]

Answer:

5873 dollars is answer pls mark as brainliest answer

8 0

3 years ago

at a flower shop, Betsy paid $33 for 3 roses and 9 tulips. Bert paid $48 for 12 roses and 8 tulips. Find the cost of one rose an

Wewaii [24]

Tulips are 3 dollars and roses are 2 dollars. Solve it using systems of equations.

6 0

3 years ago

What is 57 divided by 90?

stiv31 [10]

Answer:

0.63 should be your answer

5 0

3 years ago

Read 2 more answers

You park in a garage 3 floors below ground level. Then you get in the elevator and go up 12 floors. What floor do you get off th

Gemiola [76]

Answer: (B) 6 floors

Step-by-step explanation: Use the equation -3 + 9 to get the amount of floors above ground level.

-3 + 9 = 6

6 floors.

If this helps may I please have a brainliest.

5 0

3 years ago

Read 2 more answers

A container manufacturing company was contracted to design and manufacture cylindrical cans for fruit juice. The volume of each

gtnhenbr [62]

The volume of the cylinder is the amount of fruit juice it can contain.

The relationship between the volume and the surface area is:

$\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}$

The given parameter is:

$\mathbf{V = 0.946}$

The volume of a cylinder is calculated as:

$\mathbf{V = \pi r^2 h}$

Make h the subject

$\mathbf{h = \frac{V}{ \pi r^2}}$

The surface area (A) of a cylinder is:

$\mathbf{A = \pi r^2 + \pi rh}$

Substitute $\mathbf{h = \frac{V}{ \pi r^2}}$

$\mathbf{A = \pi r^2 + \pi r \times \frac{V}{\pi r^2}}$

$\mathbf{A = \pi r^2 + \frac{V}{r}}$

Differentiate

$\mathbf{A' = 2\pi r - Vr^{-2}}$

Set to 0

$\mathbf{2\pi r - Vr^{-2} = 0}$

Rewrite as:

$\mathbf{ 2\pi r = Vr^{-2}}$

Multiply through by r^2

$\mathbf{ 2\pi r^3 = V}$

Solve for r

$\mathbf{r = \sqrt[3]{\frac{V}{2\pi}}}$

$\mathbf{A = \pi r^2 + \frac{0.946}{r}}$

So, we have:

$\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}$

Hence, the relationship between the volume and the surface area is:

$\mathbf{A = \pi (\sqrt[3]{\frac{V}{2\pi}})^2 + \frac{0.946}{(\sqrt[3]{\frac{V}{2\pi}})}}$

Read more about surface areas and volumes at:

brainly.com/question/3628550

8 0

2 years ago

Read 2 more answers