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3 years ago

Is this a linear function?

Mathematics

1 answer:

docker41 [41]3 years ago

4 0

Answer:

Yes because it is evenly and correctly put out on the table with a slope of 2

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Find four consecutive integers whose sum is 114

bija089 [108]

Answer:

the sum of 27,28,29,30 is 114

Step-by-step explanation:

6 0

3 years ago

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Which graph best represents the solution set to this system of inequalities?

Tomtit [17]

Answer:

(3/2, -1/2)

Step-by-step explanation:

add the 2 inequalities together.

you get 2X<3 so x<3/2. plug X back in and solve for y and you get y= -1/2

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2 years ago

What is the discontinuity and zero of the function

lukranit [14]

The discontinuity are the point wherein the function is not defined.
The zeros are the points wherein f(x)=0.
Computing the discontinuity points:
Set $x-1=0$ then the discontinuity point is at $x=1$ .
Comptine the zeroes:
Set $3x^2+x-4=0$
Compute the discriminant: $1^2-4(3)(-4)=49=7^2.$
Then with quadratic formula we get the solutions:
$(-1-7)/6=-8/6\\(-1+7)/6=1$
The two zeros are $\dfrac{-8}{6},1$

7 0

3 years ago

10 divided by 4 2/3 <br> uh pls help

jok3333 [9.3K]

Answer:

2.14 or 214/100

Step-by-step explanation:

8 0

3 years ago

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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