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3 years ago

Help plz i will give brainlist:)

Mathematics

2 answers:

Virty [35]3 years ago

4 0

6:45AM
7:45 minus 15 is 7:30
7:30 minus 46 mins is 6:45
So, therefore 6:45AM :)

ICE Princess25 [194]3 years ago

4 0

She needs to leave at about 7:30
She needs to eat and get ready at about 6:55
So she should wake up at about 6:55

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A family of four goes to a salon for haircuts the cost of each haircut is $13 use distributive property to find the product 4*13

atroni [7]

4 X 13 thats all you have to do mark me as brainliest if you could

7 0

3 years ago

Read 2 more answers

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require

Aleksandr-060686 [28]

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

Step-by-step explanation:

<u>Optimizing With Derivatives </u>

The procedure to optimize a function (find its maximum or minimum) consists in :

Produce a function which depends on only one variable
Compute the first derivative and set it equal to 0
Find the values for the variable, called critical points
Compute the second derivative
Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum

We know a cylinder has a volume of 4 $ft^3$ . The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

$\displaystyle A''=2\pi+\frac{16}{r^3}$