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Yuliya22 [10]
3 years ago
13

What are the names of three collinear points

Mathematics
1 answer:
zepelin [54]3 years ago
3 0
If three points were A, B and C are collinear points and B is between A and C, then AB+BC=AC
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Convert 5 1/8 to a fractional notation
nataly862011 [7]
Basically speaking, a fractional notation is the same as a fraction itself. Thus, the best way to express the given problem in here would be to turn it into a fraction. Since the given variable is a mixed number, meaning that it is a whole number and a fraction combined, it can be turned into just a fraction or more specifically, an improper fraction. To do this, we have to take the denominator and multiply it to the whole number. After that, the product would be added to the numerator. The sum would be the new numerator and the denominator would remain the same. Thus, the answer would be 41/8
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3 years ago
What is the next letter b a d e h g j k?
MArishka [77]
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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
3 years ago
A group of students are going on a trip and the trip costs $36.50 per student which includes $9.50 for lunch, an entrance ticket
Gnesinka [82]

Answer:

The answer is D) $13.50

Step-by-step explanation:

So first what you're gonna do it subtract $9.50 from $36.50 which will get you to $27.00.

Then you'll have to figure out what half of $27.00 is.

That will get you to the answer of D) $13.50

5 0
3 years ago
What is (-i)3?<br> A. i<br> B. -1<br> C. -1<br> D. 1
alexdok [17]

Answer:

you don't have -i? because it should be -i

3 0
3 years ago
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