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Liono4ka [1.6K]
3 years ago
13

Helpppppp pl for 5 point​

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

3^{-2} * 3^3

Step-by-step explanation:

Multiply 1/9 and 27 to get 27/9 which reduces to 3. When doing these equations, you add up the exponents. To get 3, you would have to have 3^1. The get this, the only equation that would work is 3^-2 x 3^3

-2+3 gets 1. So 3^1 which gets you 3. So this is the answer.

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Hey therapy was conducted to see how many people ride a bus to school every day. 160 people reported that they ride the bus to s
xeze [42]

Answer:

200 people

Step-by-step explanation:

Let: the number of people who surveyed be: x

Therefore, (160/x) x 100=80%

=> x=(160x100)/80

=>x=16000/80

=>x= 200

Thus, 200 people surveyed.

Mark as brainliest

4 0
3 years ago
2) PLS HELP ASAP!!!
Oksi-84 [34.3K]

Answer:

Last one:  Symmetric with respect to the y-axis

Step-by-step explanation:

About the Y-AXIS

because  think about using a few points  x = -2, x = -1 , x = 1, x = 2

notice that  2*(-2)^4  =  2 *(2)^4

and 2*(-1)^4 = 2*(1)^4

8 0
3 years ago
Solve:<br> 4 - 10x &gt;-21<br> and aranh the
Luden [163]

Answer:

x < \frac{5}{2}

Step-by-step explanation:

4 - 10x > -21

-10x > -25\\\\x > \frac{-25}{-10} \\\\x > \frac{5}{2}

8 0
3 years ago
Which numbers are irrational?
irakobra [83]

Answer:

√14 and √42. hope this helps.

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
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