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Deffense [45]
3 years ago
15

What is the slope of a line perpendicular to the line whose equation is x+y=−1. Fully reduce your answer.

Mathematics
1 answer:
Leona [35]3 years ago
6 0
Make y = mx = b
y = -x - 1
Slope is -1
Perpendicular = opposite sign and reciprocal slope
1/1 = 1, slope would be 1
Solution: slope = 1
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The vertex of the parabola below is at the point (2, 4), and the point (3, 6) is on the parabola. What is the equation of the pa
Liula [17]
<span>C and B are *candidates* for being the correct solution
because f(2) = 4 for both.
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A is the correct solution because f(3)=6
4 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 ramp leads up to a building. The top of the ramp is 6 feet above the ground, and the bottom is 16 feet. What is the length of
andre [41]
We could use the Pythagorean theorem for this kind of problem I think:
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36 + 256 = C^2
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Sorry if it’s wrong and glad I could help if it’s right ❤️❤️ Take care!
7 0
3 years ago
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Amiraneli [1.4K]
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4 0
3 years ago
Which are true of the function f(x)=49(1/7)^x
Ira Lisetskai [31]

The Option (1),(4) and (5) are correct.

Step-by-step explanation:

The  function given to us  is : y= f(x) =49(1/7)^x

The following  statements  are true regarding the above function is

  • As you can see  that  the value of y(49)is defined for  all values of x , so the Domain is the set of all values of x , which is set of all real numbers.
  • X is defined for all values of y. So the Range of  all the values that y can take  is also set of all real numbers greater than zero, i.e y>0.
  • As x increased by 1, each y-value is one seventh of the previous y-value

8 0
3 years ago
Read 2 more answers
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