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Studentka2010 [4]

3 years ago

11

Write a few sentences to end Sarah’s story. How many shark teeth did she find? How many shark teeth did John find? How much mone

y did they earn?

2 answers:

timofeeve [1]3 years ago

4 0

Answer:

4, 69, $0

Step-by-step explanation:

ankoles [38]3 years ago

4 0

There is no picture?

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Complete the equation of the line through (-9,-9)(−9,−9)left parenthesis, minus, 9, comma, minus, 9, right parenthesis and (-6,0

saw5 [17]

Answer:

The equation of the line is;

y = 3x + 18

Step-by-step explanation:

We want to write the equation of the line through (-9,-9) and (-6,0)

we start by calculating the slope of the line

m = (y2-y1)/(x2-x1) = (0+ 9)/(-6+9) = 9/3 = 3

The general equation of the line is;

y = mx + c

y = 3x + c

To get c, we use any of the points

we substitute for example -6 for x and 0 for y

0 = 3(-6) + c

c = 0 + 18 = 18

So the equation of the line is;

y = 3x + 18

4 0

3 years ago

Quotes:<br> If you're cooler than me doesn't that me hotter than you

Elden [556K]

Answer:

Thats good!

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is equivalent to 4x-2=38

Tom [10]

2x4 is the answer to 4x-2=38

7 0

3 years ago

Hello please help me<br>(this is 6th grade math) <br>

user100 [1]

1:4
2:8
3:12
4:16
5:20 then just graph it according to the x and y coordinates

5 0

3 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

3 years ago