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3 years ago

There were 128 children signed up for baseball camp. If each team is made up of 12

Mathematics

2 answers:

faltersainse [42]3 years ago

6 0

Answer:

There will be 10 teams and 8 children left over.

Step-by-step explanation:

12 * 10 = 120. This is the closest you can get to 128 without going over, so that means there will be 10 teams of 12 children. 128 - 120 = 8. Therefore, there are 8 children left over.

DaniilM [7]3 years ago

6 0

Answer: There will be 10 teams with 8 kids left over.

Step-by-step explanation:

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3 years ago

Parallel to y = -x -5; through (4,-3)

eduard

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Step-by-step explanation:

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3 years ago

Read 2 more answers

The domain of this function is {-12,-6,3,15}. Y=-2/3x+7

Evgen [1.6K]

Considering the given function, the table is completed as follows:

x|y

-12|15

-6| 11

3|5

15|-3

<h3>Which is the function?</h3>

The function is defined by:

$y = -\frac{2}{3}x + 7$

Hence, when x = -12:

$y = -\frac{2}{3}(-12) + 7 = 15$

When x = -6:

$y = -\frac{2}{3}(-6) + 7 = 11$

When x = 3:

$y = -\frac{2}{3}(3) + 7 = 5$

When x = 15:

$y = -\frac{2}{3}(15) + 7 = -3$

More can be learned about functions at brainly.com/question/25537936

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6 0

2 years ago

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

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