Both the domain and range of the graph shown above is all real numbers. true or false

garri49 [273]

For both of the contests, the variable is the number of competitors which we can set as x.
Consequently, the equation should be 5.50x+34.60=4.25x+64.60

5.50x=4.25x+30

1.25x=30

x=24

The fee of the contests is the same when there are 24 competitors

5 0

3 years ago

PLEASE HELP WITH THIS

tankabanditka [31]

Answer:

Step-by-step explanation:

i Don't now how to explain the answer

8 0

2 years ago

Equation B: 3x+2x+6=6x-5 ( does that have one solution, no solution, or many solutions?) and explain why

Hoochie [10]

It has one solution.

please see the attached picture for full solution

Hope it helps

Good luck on your assignment...

7 0

3 years ago

Read 2 more answers

Can someone please help me complete the table

Lady bird [3.3K]

Answer:

Step-by-step explanation:

-2, 1, 2, 4

7 0

3 years ago

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

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

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for h:

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

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

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for h.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$

In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

7 0

3 years ago