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3 years ago

8

Explain what is a SYSTEM OF LINEAR EQUATIONS?

1 answer:

lions [1.4K]3 years ago

5 0

Is just a set of two or more linear equations. In two variables (x and y) , the graph of a system of two equations is a pair of lines in the plane.

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I need help with this!

ruslelena [56]

|12| ÷ 2 • |-5|
= 12 ÷ 2 • 5
= 6 • 5
= 30
Absolute values just make the number positive if it is negative. The number stays positive if it’s positive.

7 0

3 years ago

Find the slope of this graph, thanks

Anarel [89]

Slope = (change in 'y') / (change in 'x')

If you find two points where the line goes through the corners of boxes,
then count up the 'y' boxes and the 'x' boxes between those two points,
you'll find that the line rises 3 y-boxes for each x-box.

The slope of the line is 3 .

7 0

3 years ago

The population of a small town in Alabama is 382 people. If the growth rate is 0.7%, what will its population be in 30 years?

goldenfox [79]

Formula for growth:

y = initial • (percent growth + 1)^time/frequency of growth

So, since initial population is 382

Percent growth is 0.007

The time is 30 years

And it goes it by .07% every 1 year,

y = 382 • (1.007)^30/1

y= 470.9203732 people

Since we can’t have 0.9 of a person, the population after 30 years will be 470 people.

4 0

2 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

<em>Step(i)</em>:-

Given differential problem

y′+3 y=9 t

<em>Take the Laplace transform of both sides of the differential equation</em>

L( y′+3 y) = L(9 t)

<em>Using Formula Transform of derivatives</em>

<em> L(y¹(t)) = s y⁻(s)-y(0)</em>

<em> By using Laplace transform formula</em>

<em> </em> $L(t) = \frac{1}{S^{2} }$ <em> </em>

<em>Step(ii):-</em>

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

<em>Step(iii</em>):-

<em>By using partial fractions , we get</em>

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

<em> B = 9/3 = 3</em>

Put s = -3 in equation(i)

9 = C(-3)²

<em>C = 1</em>

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

<em> 0 = A + C</em>

<em>put C=1 , becomes A = -1</em>

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

<u><em>Step(iv):-</em></u>

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

<em>By using inverse Laplace transform</em>

<em></em> $L^{-1} (\frac{1}{s} ) =1$ <em></em>

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

<u><em>Final answer</em></u>:-

<em>Now the solution , we get</em>

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago

Please answer question number four thank you!

olga55 [171]

26( w times w times w less than nvm its less than

6 0

3 years ago