\frac{(2_1-4_{2}) ^{4} } {2} <br> Quick I need help<br> Plz

Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

goblinko [34]

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

$\sqrt{4\cdot 6+25}=2\cdot 1+C$

$C=5$

Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

or

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

$y(x)=x^2+5x$

Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

4 0

3 years ago

Average rate of change from G from x=1 to x=4 is

Maurinko [17]

Answer:

3

Step-by-step explanation:

minus the variable, 4-1 is 3.

3 0

3 years ago

Find the value of f(4) for the function.<br><br> f(a) = 2(a + 4)-4

mixer [17]

Step-by-step explanation:

4a=2a+4

C.L.T

4a-2a=4

2a=4

=2

8 0

3 years ago

Choose the correct answer below. A. Increasing the level of confidence has no effect on the interval. B. Increasing the level of

murzikaleks [220]

Answer:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

Step-by-step explanation:

Let's assume that we have a parameter of interest $\mu$ who represent for example the true mean for a population. And we can construct a confidence interval in order to estimate this parameter if we know the distribution for the statistic let's say $\bar X$ and for this particular example the confidence interval is given by:

$\bar X \pm ME$

Where ME represent the margin of error for the estimation and this margin of error is given by:

$ME = t_{\alpha/2} \frac{s}{\sqrt{n}}$

And the width for the confidence interval is given by:

$Width =2ME=2t_{\alpha/2} \frac{s}{\sqrt{n}}$

And we want to see the effect if we increase the confidence level for a interval. On this case if we increase the confidence level then the critical value for the confidence interval $t_{\alpha/2}$ would be higher and then the width of the interval would increase. So then the best answer for this case would be:

B. Increasing the level of confidence widens the interval.

3 0

3 years ago

What is 763,400 rounded to the nearest ten thousand

hichkok12 [17]

I would say 760,000

8 0

3 years ago

Read 2 more answers

\frac{(2_1-4_{2}) ^{4} } {2} Quick I need help Plz