All categories

3 years ago

Can someone please help me with my hw 20 points?

Mathematics

2 answers:

Neporo4naja [7]3 years ago

7 0

Answer:

The first equation; x-2y=8

Step-by-step explanation:

Hi there!

We're told that Ty wants to isolate x in one of the equations. To do so in either, he will need to use inverse operations to cancel out values and leave just x remaining on one side of the equation.

In the second equation, he would need to subtract both sides by 6y and then divide both sides by 4 to isolate x. It's a two-step process.

However, in the first equation, he only needs to add 2y to both sides to isolate x.

I hope this helps!

patriot [66]3 years ago

6 0

Answer:

using the first equation

cause Being that the first equation has the simplest coefficients (1, -2, for x, and y respectively), it seems logical to use it to develop a definition of one variable in terms of the other

You might be interested in

A. Find the amplitude.

Feliz [49]

Answers:

a) Amplitude = 2
b) Period = pi
c) Vertical shift = -2, which means it has been shifted down 2 units.
d) Horizontal shift = 3pi/8, this shifting is to the right.
e) There is <u> one </u> cycle between 0 and 2pi.
f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2

========================================================

Explanations:

Part (a)

The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.

Note: Amplitude is always positive as it measures a distance.

---------------------

Part (b)

For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.

The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8

So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi

The graph repeats itself every pi units, so the period is pi.

---------------------

Part (c)

The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.

---------------------

Part (d)

This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.

The parent sine function y = sin(x) goes through the origin (0,0)

However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)

This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2

This must mean we need to shift the sine graph 3pi/8 units to the right.

---------------------

Part (e)

Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).

So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.

Overall, one full cycle is between 0 and 2pi.

---------------------

Part (f)

Again I'm going to use sine instead of cosine. Refer back to part (d).

The general sine curve equation is

y = A*sin(B(x-C))+D

where

|A| = amplitude
B handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/T
C = horizontal phase shift
D = vertical shift, and ties together with the midline equation

In this case, we found that

|A| = 2
T = pi leads to B = 2pi/T = 2pi/pi = 2
C = 3pi/8
D = -2

So,

y = A*sin(B(x-C))+D

will update to

y = 2*sin(2(x-3pi/8))-2

which is one way to express the equation of the red curve. Optionally you can distribute the 2 through to (x-3pi/8).

6 0

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:

<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

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

We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.

The minimum area is

$\displaystyle A=\pi(1.084)^2+\frac{8}{1.084}$

$\boxed{ A=11.07\ ft^2}$

8 0

3 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

3 years ago

Boys to girls ratio is 2 to 3. There are 18 girls. What is total number of students

alexdok [17]

$\frac{2}{3}=\frac{boys}{18}$

3*boys=2*18

3*boys=36

boys=12

12+18=30

total number of students: 30

8 0

3 years ago

Read 2 more answers

Which vector best describes the translation below?

Nimfa-mama [501]

<h3>Answer: Choice A) <9,0></h3>

Explanation:

Focus on one of the points in the figure on the left. Let's say we go for the upper left corner point (-7, 4)

Notice it moves to the corresponding image point (2,4). It has shifted 9 units to the right to follow the translation rule $(x,y) \to (x+9, y)$ . We've added 9 to the x coordinate, and the y coordinate stays the same.

This notation can be shortened to <9, 0>

In general, the notation $(x,y) \to (x+a, y+b)$ is shortened to the translation vector notation $< a, b >$ . In this case, a = 9 and b = 0.

8 0

2 years ago