Find the difference. Write your answer in scientific notation.

Mathematics

2 answers:

loris [4]2 years ago

5 0

Answer:

147.2 × 10 raised to the 6th power.

Step-by-step explanation:

First you have to convert them normal numbers. The first one is 46,000. The second one is 3,200. The four and three help to indicate the number of zeros behind that 1. 4.6 times 10000 would be 46000. (move the decimal to the right).

Then after multiplying the numbers you have to convert them back to scientific notation. Write the number out, 147,200,000. Then slice the decimal right in between 7 and 2 to form a decimal with only a tenths place. Count how many places there are to the end, and that will be the exponent you will use.

To check answer convert to number.

Bad White [126]2 years ago

5 0

Answer:

42800

Step-by-step explanation:

First do the exponent in the first parenthesis which would get you 10000 then multiply with 4.6. Now you have 46000-(3.2· 10³)

then do the exponent in it. Now you have 3.2·1000= 3200

So now your final equation is 46000-3200

42800

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a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$