Ask question

Ask question

All categories

3 years ago

15

Consider a local community of a rock chucks in Wyoming that are increasing at a rate of 11% per month. Using the Rule of 70, ans

were the following:How many months would it take to see the rock chuck population quadruple?

1 answer:

natka813 [3]3 years ago

8 0

I am not good at math or I would help :/

You might be interested in

Select the correct answer.

sergejj [24]

G(x)/f(x) will be simplified to (x+3)(x-3)/2-x^1/2,
which will give you [0,4) ∪(4, ∞).
Choice B

5 0

2 years ago

Rewrite the parametric equations by eliminating the parameter. Select ONE.

Vinil7 [7]

Answer:

$Y=(X-7)/2$

Step-by-step explanation:

we have

$X=2t+3$ -----> equation A

$Y=t-2$ -----> $t=Y+2$ -----> equation B

Substitute equation B in equation A

$X=2(Y+2)+3$

$X=2Y+4+3$

$X=2Y+7$

solve for Y

$2Y=X-7$

$Y=(X-7)/2$

6 0

2 years ago

How many ms in a second?

professor190 [17]

1000. its making me type more stuff

4 0

1 year ago

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago

HEEEEELP which one is correct?

Nostrana [21]

Answer:

Substitution

3 0

2 years ago

Read 2 more answers