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faltersainse [42]
3 years ago
7

What is the prime factorization for 180?

Mathematics
1 answer:
Svetlanka [38]3 years ago
8 0
180
2 * 90
2 * 3 * 30
2 * 3 * 2 * 15
2 * 3 * 2 * 3 * 5
2^2 * 3^2 * 5 <===
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On May 1st, Jay’s mom gives him 1 cent. Each day, she pays him double the amount she paid the day before. How much money did Mik
crimeas [40]

Answer: $1.83

Step-by-step explanation:

1. Solve this problem: 1+2+4+6+8+10+12+14+16+18+20+22+24+26. There are 15 numbers for 15 days. You want to solve this because each day is doubled. For example: first day is 1 cent, second day is 2 cents, third day is 4 cents, and so on.

2. Answer is 183 cents.

3. Convert 183 cents to dollars.

4. Your answer is $1.83.

Hope it helps!

5 0
3 years ago
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-10 is equal to 5 subtracted from a number
kumpel [21]

-10 is equal to 5 subtracted from a number

Assume that the number is x

So the equation will be

-10=x-5

we put x at first because in subtraction the number after the word "from" comes first

Now let us find x

We need to move 5 to the left side to find x, so add 5 to both sides

\begin{gathered} -10+5=x-5+5 \\ -5=x \end{gathered}

The number is

7 0
1 year ago
I don't understand this!
12345 [234]
I csn help for the first one can't help with the colunms the answer is 1,396.
3 0
3 years ago
What is 78 divided by 19
Maslowich

Answer:

4.1

Step-by-step explanation:

7 0
2 years ago
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The The Laplace Transform of a function , which is defined for all , is denoted by and is defined by the improper integral , as
guapka [62]

Answer:

a. L{t} = 1/s² b. L{1} = 1/s

Step-by-step explanation:

Here is the complete question

The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0

Solution

a. L{t}

L{t} = ∫₀⁰⁰e^{-st}t

Integrating by parts  ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = e^{-st} and v = \frac{e^{-st}}{-s} and du/dt = dt/dt = 1

So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w

So,  ∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + \frac{1}{s} [\frac{e^{-st} }{-s}]₀⁰⁰

= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]

= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]

= -1/s[(0 - 0] - 1/s²[0 - 1]

= -1/s[(0] - 1/s²[- 1]

= 0 + 1/s²

= 1/s²

L{t} = 1/s²

b. L{1}

L{1} = ∫₀⁰⁰e^{-st}1

= [\frac{e^{-st} }{-s}]₀⁰⁰

= -1/s[exp(-∞s) - exp(-0s)]

= -1/s[exp(-∞) - exp(-0)]

= -1/s[0 - 1]

= -1/s(-1)

= 1/s

L{1} = 1/s

6 0
3 years ago
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