5.2cm second piece is 1.5 times as long

Please help, It's 7:54 and I'm tired.

RideAnS [48]

Answer:

I am pretty sure we are all tired but here is your answer 195

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the solution to this equation x -9=5

lukranit [14]

Answer:

x = 14

Step-by-step explanation:

x- 9 = 5

Note the equal sign, what you do to one side, you do to the other. Isolate the variable, x. Add 9 to both sides of the equation:

x - 9 (+9) = 5 (+9)

x = 5 + 9

x = 14

x = 14 is your answer.

~

7 0

3 years ago

If a boy has 4 pairs of shorts, 8 shirts, and 2 pairs of shoes, how many different outfits could he wear?

sveta [45]

Answer:

64

Step-by-step explanation:

I paired the first shirt with all of the pants and both shoes which was 8 outfits, but here are 8 other shirts which will also make 8 outfits.

4 0

3 years ago

Question 1 (10 points)

Irina18 [472]

Answer:

Q1 d, Q2 c, Q3 d

Step-by-step explanation:

Q1

g(x)=-3x+1

g(x)=16 means that

-3x+1=16 subtract 1 from both sides

-3x=16-1 combine like terms and divide both sides by -3

x=-15/3=-5

Q2

g(x)=3x²+4x-1

g(2) means that x=2 so substitute

g(2)=3*2²+4*2-1=12+8-1=19

Q3

domain are the x values

range are the y values

8 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

2 years ago