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galben [10]
2 years ago
9

Rick traveled 23 of the way from his parents' house to his college dorm. He has driven 294 miles. What is the total distance bet

ween his parents' house and his college dorm?
Mathematics
1 answer:
dimulka [17.4K]2 years ago
6 0

Answer:

441 miles

Step-by-step explanation:

Rick traveled \frac{2}{3} of the way from his parents' house to his college dorm. He has driven 294 miles.

This means that

\dfrac{2}{3} \text{ of the way is } 294 \text{ miles}

Then

\dfrac{1}{3} \text{ of the way is } 294\div 2=147 \text{ miles}

and

\dfrac{3}{3} \text{ of the way is } 294+147=441 \text{ miles}

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Step-by-step explanation:

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How to find 62%of $400​
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Answer:

248

Step-by-step explanation:

Solution for What is 400 percent of 62:

400 percent *62 =

(400:100)*62 =

(400*62):100 =

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Now we have: 400 percent of 62 = 248

Question: What is 400 percent of 62?

Percentage solution with steps:

Step 1: Our output value is 62.

Step 2: We represent the unknown value with $x$.

Step 3: From step 1 above,$62=100\%.

Step 4: Similarly, x=400\%.

Step 5: This results in a pair of simple equations:

62=100\%(1).

x=400\%(2).

Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both

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\frac{62}{x}=\frac{100\%}{400\%}

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5 0
3 years ago
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) 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

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\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

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A flat object with two sides, one colored red (R), the other green (G), is tossed 2 times
kompoz [17]

Answer:

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Step-by-step explanation:

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7 0
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