I need help please!!!!!

Exponential Growth and Decay The expressions below represent either exponential growth or exponential decay. Check all the expre

Tomtit [17]

Answer:

First one: 5x3(0.75)2t

and fourth one: 0.05x500(0.005)t

Step-by-step explanation:

In order to have an exponencial decay, the rate of the exponencial function needs to be lesser than 1.

The rate is the value between parenthesis, so in the first equation, the rate is 0.75, so this is an exponencial decay.

The second equation has rate = 2.4, so this is not an exponencial decay.

The third equation has rate = 1.04, so this is not an exponencial decay.

The fourth equation has rate = 0.005, so this is an exponencial decay.

5 0

2 years ago

Read 2 more answers

rishi spent 3/4 of an hour each day for 2 days kyle spent 1/4 of an hour each day for 6 days. who spent more time

myrzilka [38]

To find the answer,we need to first find out their hours spent individually.

Rishi spent:
=2×3/4
=6/4hours

Kyle spent
=6×1/4
=6/4hours

As they both spent 6/4 hours,they spent equal time.

Hope it helps!

5 0

2 years ago

HELP ME PLEASE !!!!!!!!

Olenka [21]

Answer:35

Step-by-step explanation:

if you added 85,70, and 50 you would get 205 and that is more than the number of students that were interviewed so 35 would be the correct answer

3 0

2 years ago

Read 2 more answers

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

3 years ago

Find the volume of the prism below.

WINSTONCH [101]

92 ok i cant explain

6 0

2 years ago