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3 years ago

What is the equation of this graphed line?

Mathematics

2 answers:

MrMuchimi3 years ago

6 0

We can find the slow by using the formula y2-y1/x2-x1. We have (-4, -6) and (2, 6). 6+6/2+4 is 12/6 which is 2. We need to put this into point slope form before putting this into slope-intercept form.

I will use the coordinates (2, 6).

y - 6 = 2(x - 2)

This gives us y - 6 = 2x - 4. We need to isolate y. This gives us y = 2x + 2.

I decided to use point slope form because it looks like the y-intercept is 2 and something tenths. However, the line is just off.

The answer is: y = 2x + 2

xz_007 [3.2K]3 years ago

3 0

Answer is highlighted in blue. If you have any questions feel free to ask.

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3 years ago

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What percent of 88 is 72.6? ...?

Feliz [49]

Percentage problems are simple division problems. Here, what percentage of x is y means the answer is y divided by x (times 100). Let's apply it to this question.
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3 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:

Optimizing With Derivatives

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

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