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3 years ago

The mapping shows a relationship between input and

Mathematics

1 answer:

Alborosie3 years ago

6 0

a function can't have one input mapped to multiple outputs. this is the case for where 4 maps to -2 and 2, so remove (4, -2)

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A movie theater sells adult tickets for $4 and children tickets for $2. At a certain movie, there were 90 people in attendance a

Juli2301 [7.4K]

[X,Y]=40,50. 40 4$ tickets and 50 2$ tickets.

8 0

2 years ago

The equation x2 16 + y2 9 = 1 defines an ellipse, which is graphed above. in this excercise we will approximate the area of this

Citrus2011 [14]

(a) 4
(b) y = sqrt(9 - (9/16)x^2)
The best guess to the formula using knowledge of the general formula for an ellipse is:
x^2/16 + y^2/9 = 1
(a). An ellipse is reflectively symmetrical across both the major and minor axis. So if you can get the area of the ellipse in a quadrant, then multiplying that area by 4 would give the total area of the ellipse. So the factor of 4 is correct.
(b). The general equation for an ellipse is not suitable for a general function since it returns 2 y values for every x value. But if we restrict ourselves to just the positive value of a square root, that problem is easy to solve. So let's do so:
x^2/16 + y^2/9 = 1
x^2/16 + y^2/9 - 1 = 0
x^2/16 - 1 = - y^2/9
-(9/16)x^2 + 9 = y^2
9 - (9/16)x^2 = y^2
sqrt(9 - (9/16)x^2) = y
y = sqrt(9 - (9/16)x^2)

4 0

3 years ago

The area of the following rectangle is 30 square units. What is the value of x?

Pepsi [2]

Hello!

To find the area of a rectangle you do length * width

You can plug in the values you know

(x - 2) * 5 = 30

Divide both sides by 5

x - 2 = 6

Add 2 to both sides

x = 8

The answer is 8

Hope this helps!

4 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

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