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3 years ago

2 points

Mathematics

1 answer:

miskamm [114]3 years ago

5 0

The x - intercept of 5x - 3y = 15 is (3, 0)

The y -intercept of 5x - 3y = 15 is (0, -5)

<h3>Solution:</h3>

Given equation is 5x - 3y = 15

To find: x - intercept and y -intercept

The x intercept is the point where the line crosses the x axis. At this point y = 0

The y intercept is the point where the line crosses the y axis. At this point x = 0.

Finding x - intercept:

To find the x intercept using the equation of the line, plug in 0 for the y variable and solve for x

So put y = 0 in given equation

5x - 3(0) = 15

5x = 15

x = 3

So the x - intercept is (3, 0)

Finding y - intercept:

To find the y intercept using the equation of the line, plug in 0 for the x variable and solve for y

So put x = 0 in given equation

5(0) - 3y = 15

-3y = 15

y = -5

So the y - intercept is (0, -5)

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3 years ago

I need help solving this equation 4n-9=2(5+2n)

Nastasia [14]

First, use distributive property on the right half.

2 * 5 = 10

2 * 2n = 4n

4n - 9 = 10 + 4n

Add 9 to both sides

4n = 19 + 4n

Subtract 4n from both sides

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3 years ago

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Snowcat [4.5K]

Answers with Method:-

1) Multiply the length value by 100,000.

$2.5 \: hm = 2.5 \times 100000$

$= >2.5 \: hm = 250000 \: mm$

2) For this, divide the length value by 1000.

$= > \frac{1800}{1000}$

$= > 1.8 \: dam$

3) For finding the approximate value, just multiply the value of length by 1.609.

$= >6450 \: m = (6450 \times 1.609) \: km$

$= > 10380.27 \: km$

4) Multiply the given mass value by 100.

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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:

Explanation:-

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

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

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

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

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

Step(iii):-

By using partial fractions , we get

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

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

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)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

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)

0 = A + C

put C=1 , becomes A = -1

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

Step(iv):-

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

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

Applying inverse Laplace transform on both sides

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

By using inverse Laplace transform

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

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

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

Final answer:-

Now the solution , we get

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

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3 years ago

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sladkih [1.3K]

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