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3 years ago

Evan can text 96 words in 12 minutes. At this rate, how many words could he text in

1 answer:

7 0

If you divide 96 by 12 you get 8. This means he texts 8 words per minute. If you then multiply 8 by 14 you get 112 which means he can text 112 words in 14 minutes.

You might be interested in

What is the sum 10 + (-10)

Marta_Voda [28]

Answer:

Step-by-step explanation:

because

k c r

10 + (-10)

10 - 10=0

8 0

3 years ago

Read 2 more answers

Can you help me please?

PilotLPTM [1.2K]

Answer:

Your answer should be 0.125

Step-by-step explanation:

Divide the numerator by the denominator

1 divided by 8 = 0.125

I see you're from k12. Hello, fellow homeschooler. Good luck on the assignment!

5 0

3 years ago

) 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

Leanne jumped a distance of 3 feet 8 inches. Monica jumped a distance of 4 feet 6 inches. How many more inches did Monica jump t

mote1985 [20]

Answer:

10 more inches

Step-by-step explanation:

6 0

2 years ago

30 POINTS!! ASAP! PLS SHOW WORK

sp2606 [1]

Answer:

QII

Step-by-step explanation:

We can use the acronym: ASTC or All Students Take Calculus.

The first letter of each word tells us details within a certain quadrant.

All trig functions in QI will be positive.

Only sine (and cosecant) will be positive in QII.

Only tangent (and cotangent) will be positive in QIII.

And only cosine (or secant) will be positive in QIV.

We know that tan(θ)<0. In other words, it's negative.

So, it our angle θ <em>cannot</em> be in QI or QIII.

We also know that cos(θ)<0.

So, this removes QIV, since in QIV, cosine is positive.

Therefore, the only choices that remains is QII.

In QII, sine is positive, tangent is negative, and cosine is negative.

This fits all our conditions, so θ is in QII.

And we're done!

3 0

3 years ago