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3 years ago

14

Ryan has shells and 4 equal groups tanner has 9 more than Ryan tanner has 3 fewer shells than parker. Parker has 48 shells. How

many shells are in each of ryan's group

1 answer:

Nataly [62]3 years ago

3 0

Answer:

9.75

Step-by-step explanation:

48-3=45

45-9=38

38÷4=9.75

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3 years ago

Identify the missing term in the following geometric sequence:

aleksley [76]

Answer:

B. -1/8

Step-by-step explanation:

Using the terms given in the sequence:

4, -2, 1, -1/2, 1/4

we can find the pattern by dividing since it is a geometric seqence:

-2/4 = -1/2

1/-2 = -1/2

-1/2 ÷ 1 = -1/2

So, each term is multiplied by -1/2 to get the next term in the sequence:

$\frac{1}{4}*-\frac{1}{2}=-\frac{1}{8}$

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3 years ago

What is the value of x

il63 [147K]

the 2 angles are congruent so...

x+40=3x set the 2 equations equal to each other

40=2x subtract x from both sides

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Have a blessed day!

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3 years ago

Read 2 more answers

(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

3 years ago

a horse is shown in the scale drawing the height of the horse in the drawing is 4.5 inches the scale that maps the drawing to th

Radda [10]

4.5 inches multiplied by 1.5 feet equals 6.75 feet.

6 0

3 years ago