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2 years ago

Its easy pls help me

Mathematics

1 answer:

AysviL [449]2 years ago

7 0

Answer:

start on point (0,2) and go up 3 over 1 and mark that point...continue with the pattern and draw the line connecting all the points

Step-by-step explanation:

when youre done it should look like this

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Work out the value of x-y

krek1111 [17]

Answer:

x=5

y=3

Step-by-step explanation:

easy question to solve

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2 years ago

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The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

3 years ago

HELP

madreJ [45]

It’s to hard to see what you need help on e

8 0

2 years ago

What is 3 to the 3rd power + 6 parentheses 2 + 3/6 parentheses equal

dedylja [7]

The answer is 20 im almost sure

6 0

2 years ago

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If (-2) = -14 for f(x) = 3x -b; find b? -

Verizon [17]

Answer:

The value of b is 8.

Step-by-step explanation:

The given function is,

$f(x) = 3x -b$

The given data is,

$f(-2)=-14$

i.e when the value of x is -2, then the value of the function is -14.

Putting all the values,

$\Rightarrow -14 = 3(-2) -b$

$\Rightarrow -14 = -6 -b$

$\Rightarrow b=-6+14$

$\Rightarrow b=8$

Therefore, the value of b is 8.

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3 years ago

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