Ask question

Ask question

All categories

3 years ago

15

If 18.9 million computers were in use in 1980 and the number is rising by 19% annually, predict the number of computers in use i

n 2015. Round to the nearest whole number

1 answer:

4vir4ik [10]3 years ago

4 0

18.9(1+0.19)^35, 35 is the number of years
use your calculator, you should get 661.5, round to the newest whole number 662million

You might be interested in

This table shows a linear relationship between the amount of water in a bath tub and time.

MissTica

Answer:

A. The rate is -2 each hour

B. It is decreasing, as the hours increase, the gallons decrease

6 0

2 years ago

Read 2 more answers

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago

Solve. Good luck! Please do not try to google this.

Elena L [17]

$\frac{x^{2} + x - 2}{6x^{2} - 3x} = \sqrt{2x} + \frac{3x^{2}}{2}$
$\frac{x^{2} + 2x - x - 2}{3x(x) - 3x(1)} = \frac{2\sqrt{2x}}{2} + \frac{3x^{2}}{2}$
$\frac{x(x) + x(2) - 1(x) - 1(2)}{3x(x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{x(x + 2) - 1(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$\frac{(x - 1)(x + 2)}{3x(2x - 1)} = \frac{2\sqrt{2x} + 3x^{2}}{2}$
$3x(2x - 1)(2\sqrt{2x} + 3x^{2}) = 2(x + 2)(x - 1)$
$3x(2x(2\sqrt{2x} + 3x^{2}) - 1(2\sqrt{2x} + 3x^{2}) = 2(x(x - 1) + 2(x - 1))$
$3x(2x(2\sqrt{2x}) + 2x(3x^{2}) - 1(2\sqrt{2x}) - 1(3x^{2})) = 2(x(x) - x(1) + 2(x) - 2(1)$
$3x(4x\sqrt{2x} + 6x^{3} - 2\sqrt{2x} - 3x^{2}) = 2(x^{2} - x + 2x - 2)$
$3x(4x\sqrt{2x} - 2\sqrt{2x} + 6x^{3} - 3x^{2}) = 2(x^{2} + x - 2)$
$3x(4x\sqrt{2x}) - 3x(2\sqrt{2x}) + 3x(6x^{3}) - 3x(3x^{2}) = 2(x^{2}) + 2(x) - 2(2)$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} + 18x^{4} - 9x^{3} = 2x^{2} + 2x - 4$
$12x^{2}\sqrt{2x} - 6x\sqrt{2x} = -18x^{4} + 9x^{3} + 2x^{2} + 2x - 4$
$6x\sqrt{2x}(2x) - 6x\sqrt{2x}(1) = -9x^{3}(2x) - 9x^{3}(-1) + 2(x^{2}) + 2(x) - 2(2)$
$6x\sqrt{2x}(2x - 1) = -9x^{3}(2x - 1) + 2(x^{2} + x - 2)$

5 0

3 years ago

Which expressions are equivalent to the one below?check all that apply. 21^x/3^x

mafiozo [28]

C and E and F are all the correct answers

7 0

2 years ago

Read 2 more answers

Find the coordinates of the midpoint of a segment with the endpoints (16,5) & (28,-13)

frutty [35]

$\text{The formula of the midpoint of a segment with the end points:}\\\\A(x_A;\ y_A)\ \text{and}\ B(x_B;\ y_B)\\\\M_{AB}=\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)$

$\text{We have:}\\A(16;\ 5)\to x_A=16;\ y_A=5\\B(28;\ -13)\to x_B=28;\ y_B=-13\\\\\text{substitute:}\\\\M_{AB}=\left(\dfrac{16+28}{2};\ \dfrac{5+(-13)}{2}\right)=\left(\dfrac{44}{2};\ \dfrac{-8}{2}\right)=(22;\ -4)$

7 0

3 years ago