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2 years ago

When will I be happy again

Mathematics

2 answers:

DerKrebs [107]2 years ago

4 0

Answer:

never

Step-by-step explanation:

sorry but never I'm just kidding get a girl friend and you will be good

ss7ja [257]2 years ago

3 0

Answer:

Oh mna. I hope you will be happy soon man.

Step-by-step explanation:

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Kindly solve ½=4/?, What's going to be the answer? Thank you

Luda [366]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the line through (3, 1, −2) that intersects and is perpendicular to the line x = −1 + t, y = −2 + t, z = −1 + t. (HINT: If

mel-nik [20]

Answer:

( xo , yo , zo ) = ( 1 , 0 , 1 )

Step-by-step explanation:

Given:-

- A line passing through point (3, 1, −2) intersects and is perpendicular to line with coordinates:

x = −1 + t, y = −2 + t, z = −1 + t .... t = arbitrary parameter.

Find:-

The coordinates for point of intersection.

Solution:-

- The line that passes through point (3, 1, −2) = ( a, b , c ) and an a arbitrary point on the given line have the following direction vector d2 :

d2 = ( x2 , y2 , z2 )

x2 = a - ( x ) = 3 - ( -1 + t ) = 4 - t

y2 = b - ( y ) = 1 - ( -2 + t ) = 3 - t

z2 = c - ( z ) = -2 - ( -1 + t ) = -1 - t

d2 = ( 4 - t , 3 - t , -1 - t )

- The direction vector d1 of the given line is:

d1 = ( x1 , y1 , z1 )

x1 = 1

y1 = 1

z1 = 1

d1 = ( 1 , 1 , 1 )

- The dot product of two orthogonal vectors is always equal to zero:

d1.d2 = 0

( 4 - t , 3 - t , -1 - t ) . ( 1 , 1 , 1 ) = 0

- Solve for parameter (t):

(4 - t) + (3 - t) + (-1 - t) = 0

6 -3t = 0

t = 2

- The coordinates of the point of intersections can be evaluated by substituting the value of "t" into the given equation of line:

xo ( t = 2) = - 1 + 2 = 1

yo ( t = 2) = - 2 + 2 = 0

zo ( t = 2) = - 1 + 2 = 1

- The coordinates are:

( xo , yo , zo ) = ( 1 , 0 , 1 )

8 0

3 years ago

Plz I need help with the last 2 and plz simplify then

GalinKa [24]

Answer:

c) (8^6) / 8 = 32768

d) (7^8) / (7^6) = 49

Step-by-step explanation:

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

7b^2-b+3b+6-3b^2+5+2b^2+7b

slava [35]

6b^2 + 9b + 11 is that in simplified form. I hope that it is what you were looking for :)

4 0

3 years ago

When will I be happy again​

When will I be happy again