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3 years ago

PLEASE HELP IM JUST TRYING TO MOVE FORWARD I DONT GET THESE

Mathematics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

38.7 (rounding to the nearest 10th)

Step-by-step explanation:

tanx= 8/10

tan^-1(8/10)=x

x= about 38.7

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5 divided by 60 5/60 ??

mr Goodwill [35]

Answer:

1/12 or 0.08333333333

Step-by-step explanation:

5/60 can be factored to 1/12 because the two numbers are each divisible by 5.

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3 years ago

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How do I solve this problem

CaHeK987 [17]

Use cymath it gives you the answer :)

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3 years ago

What is the sum of the first 18 terms of the arithmetic series 11+15+19+23+...?

agasfer [191]

810
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3 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

2 years ago

Solve for x Find ADB Find BDC

NeTakaya

Answer:

x=12 ADB= 56 BDC= 54

Step-by-step explanation:

ADB and BDC are complementary, so they add up to 90 degrees. In other words ADB+BDC= 90 So I can input my knowns into the equation

(3x+10) + (4x-4) = 90

I combined like terms to give me 7x+6=90

Then I subtract 6 from both sides giving me 7x=84

Last I divide 7 on both sides. x=12

Then I input the x (12) into the equation and solve it from there

3(12)+10 and 4(12)-4

3(12)+10=56

4(12)-4=54

. . . I think. Hope this helps!

5 0

3 years ago