Ask question

Ask question

All categories

3 years ago

6

A scuba diver is 72 feet below sea level .As the diver rise he stops every 15 feet what integer tells his depth after the first

pause?

1 answer:

Ksju [112]3 years ago

8 0

The diver is at -72 because he is below sea level 0. He stops every 15 ft...After the first pause, he is at -72 + 15 = - 57

answer : -57

You might be interested in

Marina spent $13.50 at the grocery store. She bought pears, kiwis, and pineapples. Pears cost $0.50 each, pineapples cost $1.50

MArishka [77]

Answer:

Number of pineapples= 10

number of pears = 19

number of kiwis = 10 -2 = 8

Step-by-step explanation:

Cost of a pear = $ 0.50

Cost of a pineapple = $ 1.5

cost of a kiwi = $ 0.3

Let the pineapples = p

Number of pears = 9 + p

Number of kiwis = p - 2

The cost is

0.15 p + 0.5 (9 + p) + 0.3 (p - 2) = 13.50

0.15 p + 4.5 + 0.5p + 0.3 p - 0.6 = 13.50

0.95 p = 9.6

p = 10

8 0

3 years ago

Read 2 more answers

A notha wun ........

geniusboy [140]

Slope is -1/5

Y-intercept is 21

Equation is y=-1/5x+21

Yes it’s proportional :)

5 0

3 years ago

Read 2 more answers

Forks come in packages of 10 and spoons come in packages of 8.jayrod wants to have the same number of forks and spoons and does

uysha [10]

5 packages of spoons and and 4 packages of forks

7 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

3 years ago

Help in 5 plz I will give brainlist

elena-14-01-66 [18.8K]

M=14.6 (is rounded to the nearest tenth.)

8 0

3 years ago