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3 years ago

2. Delia and Beto left for school at the same time.

Mathematics

1 answer:

Tema [17]3 years ago

5 0

Answer:

Beto will get to the school first.

Step-by-step explanation:

We are given that Delia lives 3/4 mile from school and walks at a pace of

16 minutes per mile.

So Delia takes $\frac{3}{4} \times 16 =$ 12 minutes to reach school.

While Beto lives 3 1/2 miles from school and bikes at a pace of 3 minutes per mile.

So Beto takes $\frac{7}{2} \times 3 =$ 10.5 minutes to reach school.

Therefore, Beto will get to the school first.

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(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

The half life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am

Veronika [31]

Using an exponential function, it is found that:

a) $N(t) = 75(0.5)^{\frac{t}{3.8}}$

b) 37.5 grams of the gas remains after 3.8 days.

c) The amount remaining will be of 10 grams after approximately 11 days.

<h3>What is an exponential function?</h3>

A decaying exponential function is modeled by:

$A(t) = A(0)(1 - r)^t$

In which:

A(0) is the initial value.

r is the decay rate, as a decimal.

Item a:

We start with 75 grams, and then work with a half-life of 3.8 days, hence the amount after t daus is given by:

$N(t) = 75(0.5)^{\frac{t}{3.8}}$

Item b:

This is N when t = 3.8, hence:

$N(t) = 75(0.5)^{\frac{3.8}{3.8}} = 37.5$

37.5 grams of the gas remains after 3.8 days.

Item c:

This is t for which N(t) = 10, hence:

$N(t) = 75(0.5)^{\frac{t}{3.8}}$

$10 = 75(0.5)^{\frac{t}{3.8}}$

$(0.5)^{\frac{t}{3.8}} = \frac{10}{75}$

$\log{(0.5)^{\frac{t}{3.8}}} = \log{\frac{10}{75}}$

$\frac{t}{3.8}\log{0.5} = \log{\frac{10}{75}}$

$t = 3.8\frac{\log{\frac{10}{75}}}{\log{0.5}}$

$t \approx 11$

The amount remaining will be of 10 grams after approximately 11 days.

More can be learned about exponential functions at brainly.com/question/25537936

4 0

2 years ago

Order: Anadrol-50 (oxymetholone) 200 mg po daily. The recomnes patient who has bipolar disorder? 3 kg 50 e ste for Is the prescr

VMariaS [17]

Answer:

Yes the ordered dose of 200 mg/d lies within the recommended range of 75 mg/d to 375 mg/d

Step-by-step explanation:

Given:

Ordered dose of Anadrol-50 (oxymetholone) = 200 mg per day

Recommended range = 1-5 mg/kg/d

Weight of the patient = 75 kg

Now,

For the patient weighing 75 kg, the recommended dose will be

Minimum dose will be

= Minimum value of Recommended range × Weight of the patient

= 1 mg/kg/d × 75 kg

= 75 mg/d

and,

the Maximum dose will be

= Maximum value of Recommended range × Weight of the patient

= 5 mg/kg/d × 75 kg

= 375 mg/d

Yes the ordered dose of 200 mg/d lies within the recommended range of 75 mg/d to 375 mg/d

4 0

4 years ago