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3 years ago

Prove that for every two positive integers a and b that

1 answer:

6 0

$$(a+b)(\frac{1}{a}+\frac{1}{b}) \ \textgreater \ or = 4$
\\
1+a/b+b/a + 1 \geq 4
$ $$(a+b)(\frac{1}{a}+\frac{1}{b}) \ \textgreater \ or = 4$
$

distributed

choose 1 as your lowest positive integer

plug 1 into a and b

1 + 1 + 1 + 1 $\geq 4$

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Olives garden is only 10 yd and the watermelon plants she wants to grown requir 2.5yd each how many watermelon plants can she gr

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Answer:

Step-by-step explanation:

10÷2.5

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3 years ago

Can someone help me (3.7x10-8)(2.9x10⁶

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Scientific notation:
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3 years ago

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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

ACTIVITY 1

aleksklad [387]

Answer:

Question A:

1. 1/2 = 0.5 (Terminating)

2. 5/6 = 0.833333.. (Repeating)

3. 21/3 = 1.6666666.. (Repeating)

Question B:

1. 0.6 = 3/5

2. 1.25 = 5/4 = 1 1/4

3. 0.125 = 1/8

Step-by-step explanation:

A. Write the fraction or mixed number as a terminating or repeating decimal.

In order to convert a fraction into decimal, the numerator has to be divided by denominator.

1. 1/2

The answer is: 0.5 which is a terminating decimal.

2. 5/6

The answer is: 0.833333.. which is a repeating decimal.

3. 2 1/3 (assuming the question is this)

$2\frac{1}{3} = \frac{7}{6} = 1.6666666..$

The answer is 1.6666666.. which is a repeating decimal.

B. Write the terminating decimal as fraction or mixed number i n simplest form.

1. 0.6

$0.6 = \frac{6}{10} = \frac{3}{5}$

2. 1.25

$1.25 = \frac{125}{100} = \frac{5}{4} = 1\frac{1}{4}$

3. 0.125

$0.125 = \frac{125}{1000} = \frac{5}{40} = \frac{1}{8}$

Hence,

Question A:

1. 1/2 = 0.5 (Terminating)

2. 5/6 = 0.833333.. (Repeating)

3. 21/3 = 1.6666666.. (Repeating)

Question B:

1. 0.6 = 3/5

2. 1.25 = 5/4 = 1 1/4

3. 0.125 = 1/8

7 0

3 years ago