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3 years ago

Using the graph f(x)=x^2 as a guide, describe the transformations, and then graph each function.

Mathematics

1 answer:

Margarita [4]3 years ago

6 0

Down refers to the movement along the y axis and left and right refer to movement along the x axis. 20. down 2, no y movement 21. 5 to the left, no y movement 22. to the right 1, no y movement 23. 4 to the left, down 3 24. to the left 2 and up 2 25. 4 to the right and down 9 26. opens up a bit wider, no x or y movement (picture it opening like a flower blooming) 27. opens downward and is very very close to the y axis 28. opened up SUPER wide, like a large shallow bowl.

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ABCD is a parallelogram. Find the measure of angle ABC.

iVinArrow [24]

Answer:

∠ABC = 50°

Step-by-step explanation:

Since this is a parallelogram, opposite angles are congruent.

Therefore, ∠D ≅ ∠B

∠D = 38° + 12°

∠B = 38° + 12°

∠B = 50°

∠ABC = 50°

5 0

3 years ago

Read 2 more answers

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

Someone help :d<br> look at the screenshot i inserted anything helps have a nice day :D

11111nata11111 [884]

Answer:

$18.24 (C)

Step-by-step explanation:

15.20 x 0.20 = 3.04

3.04 + 15.20 = 18.24

Hope this helped!

8 0

3 years ago

What is the midpoint of the segment shown below

asambeis [7]

Midpoint = $( \frac{X2+X1}{2} , \frac{Y2+Y1}{2} )$

Plug in:

$( \frac{3+7}{2} , \frac{4+(-2)}{2} )$

Equals:

$( \frac{10}{2} , \frac{2}{2})$

Your answer is then:

(5, 1), your midpoint.

3 0

3 years ago

Read 2 more answers

Given the graph below, which of the following statements is true?

iogann1982 [59]

Answer: D.

The graph does not represent a one-to-one function because the y-values between 0 and 2 are paired with multiple x-values.

Step-by-step explanation:

Looking at the graph included in the answer, the graph is connected between four points.

It starts with a negative slope, then changes to a positive slope after (-2, 0), and then changes back to a negative slope after (2, 2).

To answer this, first we need to know what a one-to-one function is.

A one-to-one function is any function that only has one x-coordinate for every y-coordinate. If a function has two x-coordinates for one y-coordinate (for example, (5, 4) and (10, 4)), then the function is not one-to-one.

An example of a one-to-one function is the graph of y = x.

There is only one x-coordinate for every y-coordinate on the graph.

<em>You can use the "horizontal line" test to determine whether a function is one-to-one.</em> If a horizontal line is placed anywhere on a graph and intersects the function more than once, the function is not one-to-one

For the function given:

When a horizontal line is placed at "y = 1", it intersects the function three times. This means that the function IS NOT one-to-one.

Choice D is correct.

The coordinates between y = 0 and y = 2 are paired with multiple x-values, and this correctly explains why the function is not one-to-one.

8 0

3 years ago