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3 years ago

Please help me! Only have about 10 minutes until I have to submit this assignment!

Mathematics

2 answers:

salantis [7]3 years ago

8 0

Answer:

626

Step-by-step explanation

irinina [24]3 years ago

4 0

The answer is 626 I think

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Solve the equation by completing the square 5x(x+6)=-50

motikmotik

First lets distribute the 5x
5x^2 + 30x = -50

Lets divide every term by 5
x^2 + 6x = -10

To complete the square we have to half the b value, which in this case is 6. Then square it.
Half of 6 is 3, 3 squared is 9

Add that to both sides of the equation
x^2 + 6x + 9 = -1

Find the binomial squared
(x+3)^2 (If you're wondering how i got that please comment)

(x+3)^2 = -1

Take the square root of the equation of both sides
(x+3) = +/- i
x = -3 +/- i
x = -3 - i
and
x = -3 + i

3 0

3 years ago

( BRAINLIEST IF CORRECT)If your business is generating $500 per month in cash flow, and you are trying to save up for a new pitc

svet-max [94.6K]

9 months it gives you $200 extra money

7 0

3 years ago

You buy magazines for $5 and books for $10. You spent a total of $100. Let x represent the number of magazines and y represent t

bagirrra123 [75]

The answer would be 10x+5y=100

8 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:

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

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

5 0

3 years ago

TQ is equal to or congruent to QR. Find m∠SQT. A. 21 B. 10 C. 4 D. 69

Law Incorporation [45]

D is the answer to this question.....

3 0

3 years ago

Read 2 more answers