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zhannawk [14.2K]

3 years ago

12

The population of Webb County, Texas, from the year 2000 through 2010 is shown in the graph.

1 answer:

olga_2 [115]3 years ago

6 0

Answer:

The is answer is B. 278,500

Step-by-step explanation:

You might be interested in

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

Recall that:

$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

$y(t) = 7 -7e^{-t} +e^{-(t-3)} u (t-3)$

$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

$\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

3 0

2 years ago

Pleaseeeee hurryy need help!!

lord [1]

it is the last one

Step-by-step explanation:

by simplifying powers it becomes

8g^8/64h ^5

simplifying it further gives

g^8/ 8h^5

8 0

3 years ago

Read 2 more answers

which of the following fractions are equivalent? 3/4 and 6/7. 2/3 and 12/13. 1/9 and 4/12. 1/6 and 3/18

sertanlavr [38]

Answer:

1/6 and 3/18

Step-by-step explanation:

I used the butterfly method to compare all the fraction pairs together and then 1/6 and 3/18 were equal. Hopefully I can help you.

8 0

2 years ago

Need help !!!wuicknkkkkkkk

amm1812

Answer:

do it

Step-by-step explanation:

can determine that 8 is more than -9 by looking which one is more further to the right, since I see 8 I assume it is 8 and that would be correct

3 0

2 years ago

How to solve if 5a=6 and 7b=8, 35ab= <br> (please explain, i need explaination)

ser-zykov [4K]

Answer:

35ab = 48

Step-by-step explanation:

given

5a = 6 ( divide both sides by 5 )

a = $\frac{6}{5}$

and

7b = 8 ( divide both sides by 7 )

b = $\frac{8}{7}$

substitute these values for a and b into the expression and simplify

35ab

= 35 × $\frac{6}{5}$ × $\frac{8}{7}$ ( divide 35 and 5 by 5 )

= 7 × 6 × $\frac{8}{7}$

= 42 × $\frac{8}{7}$ ( divide 42 and 7 by 7 )

= 6 × 8

= 48

6 0

1 year ago