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s2008m [1.1K]
3 years ago
8

HELP ASAP! GIVING BRAINLIEST

Mathematics
1 answer:
beks73 [17]3 years ago
6 0

W has 1 line of symmetry- right down the middle. Nowhere else on the figure can it be reflected into symmetry.

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The curve given by x=sin(t),y=sin(t+sin(t)) has two tangent lines at the point (x,y)=(0,0). List both of them in order of increa
SOVA2 [1]

Answer:

y = 0

y =2x

Step-by-step explanation:

Given parametric equations:

x (t) = sin (t)

y (t) = sin (t + sin (t))

The slope of the curve at any given point is given by dy / dx we will use chain rule to find dy / dx

(dy / dx) * (dx / dt) = (dy / dt)

(dy / dx) = (dy / dt) / (dx / dt)

Evaluate dx / dt and dy / dt

dx / dt = cos (t)

dy / dt = cos (t + sin (t)) * (1+cos (t))

Hence,

dy / dx = (1+cos(t))*cos(t + sin (t))) / cos (t)

@Given point (x,y) = 0 we evaluate t

0 = sin (t)

t = 0 , pi

Input two values of t and compute dy / dx

@ t = 0

dy / dx = (1 + cos (0))*cos (0 + sin (0))) / cos (0)

dy / dx = (1+1)*(1) / (1) = 2 @ t = 0

@t = pi

dy / dx = ( 1 + cos (pi))* cos (pi + sin (pi)) / cos (pi)

dy / dx = (1-1) * (-1) / (-1) = 0 @ t = pi

The corresponding gradients are 0 and 2 in increasing order and their respective equations are:

y = 2x

y = 0

5 0
2 years ago
What is the shifts of mathematics?
jasenka [17]
On photos is answer

7 0
2 years ago
10 points!!!!!?!!!! Tyrone wants to know how much wrapping paper is needed to cover a box with the dimensions shown
Leviafan [203]

Answer: B.54 Square inches

Step-by-step explanation:  Surface Area

= 2(lb + bh + lh)

= 2(6 x 3 + 3 x 1 + 6 x 1)

= 2(18 + 3 + 6)

= 2(27)

= 54 square inches

7 0
2 years ago
In a certain town 215 boys and 265 girls were born last year. What is the probability that a child chosen at random from among t
Tanya [424]

Answer:

What I think it is from 5 through 25 or something like that

5 0
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:

<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
2 years ago
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