Help please thank you!!!

In Youngtown, the population reached 4420 people in 1993. In 1998, there were 5600 residents. What was the population in the yea

Ede4ka [16]

Growth rate = (Present - Past)/Past Plugging in what we know Growth rate =(5600â’4420)/4420 Thus Growth rate=.26697 Now we can plug the growth rate into our first formula which gives us P=5600e^(.26697â‹…2) Solve for P and we get P = 9551.58 however since you can not have .58 of a person we round down to 9551. So Youngtown will have 9551 citizens in the year 2000

6 0

4 years ago

Regular hexagon ABCDEF is inscribed in a circle with center H. What is the image of segment BC after a reflection over line FC?

AlladinOne [14]

Regular hexagon ABCDEF is inscribed in a circle with center H. What is the image of segment BC after a reflection over line FC?

5 0

3 years ago

How many digits are written when 1000 to the power of 2015 is expressed as a numeral?

PtichkaEL [24]

Answer: There are 6046 digits will be written on expansion upto 2015th power.

Step-by-step explanation:

Since we have given that

$1000^{2015}$

We have to find the number of digits ,

So,

$1000^{2015}\\\\=(10^3)^{2015}\\\\=10^{6045}$

Since we know that

$10^n\text{ has n+1 number of digits }$

so,

$10^{6045}\text{ has 6046 number of digits }$

Hence, there are 6046 digits will be written on expansion upto 2015th power.

6 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:

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

After 20 minutes Juan had it completed 12 questions, which is 0.7 of his assignment. Which percent of the Simons have he not com

egoroff_w [7]

To find the incompleted portion of the test, convert the decimal to a percentage by moving the decimal two places to the right. .7 as a percentage is 70%. subtract 70% from 100%, giving you 30%. Juan has not completed 30% of his test, or 5 questions.

4 0

3 years ago