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3 years ago

10

I need to pass this one, please help (p.s) It would really help if you hovered over my account because I'm making these quick so

answering them would help

1 answer:

grin007 [14]3 years ago

6 0

Answer:

5 times greater

Step-by-step explanation:

Slope of function 1 is 10

slope of function 2 is 2

10/2=5

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What is the slope of this line? <br> A.) −3/4<br> B.) 3/4<br> C.) −4/3<br> D.) I don't know.

alina1380 [7]

A: -3/4
Rise/Run
3 is the rise
4 is the run

3 0

2 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

What are the approximate solutions of 2x^2 + x = 14, rounded to the nearest hundredth?

Slav-nsk [51]

Hello,

2x²+x-14=0
Δ=1+4*14*2=113

x=(-1-√113)/4=-2,8839...≈-2.88
or x=(-1+√113)/4=2,40753≈2.41

7 0

3 years ago

Which compares the modes of the data?

sveta [45]

Answer:

I think C

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Transversal cuts parallel lines and at points X and Y respectively. The point nearest to C is X, and the point nearest to D is Y

hammer [34]

Answer: The answer is (B) ∠SYD.

Step-by-step explanation: As mentioned in the question, two parallel lines PQ and RS are drawn in the attached figure. The transversal CD cut the lines PQ and RS at the points X and Y respectively.

We are given four angles, out of which one should be chosen which is congruent to ∠CXP.

The angles lying on opposite sides of the transversal and outside the two parallel lines are called alternate exterior angles.

For example, in the figure attached, ∠CXP, ∠SYD and ∠CXQ, ∠RYD are pairs of alternate exterior angles.

Now, the theorem of alternate exterior angles states that if the two lines are parallel having a transversal, then alternate exterior angles are congruent to each other.

Thus, we have

∠CXP ≅ ∠SYD.

So, option (B) is correct.

5 0

2 years ago