3. True or False? If false, how would you correct the statement?

How is 4^m times 4^n different from (4^m)^n?

xenn [34]

Well 4^m*4^n can be added so lets say you have 4^4*4^6 that would just be adding the exponets to get 4^10. but (4^m)^n that is multiplying the exponet of 4 to n so (4^6)^4 would be 4^24.

4 0

4 years ago

Chelsea takes the bus to school, but she walks home. The bus travels to the school at an average speed of 45 miles per hour. Whi

Neko [114]

Answer:

time walking=1-t hrs

ditace is same

therefore 2.5*(1-t)=45*t

2.5-2.5t=45t

t=2.5/46.5 hrs

this has only one solution since it is in linear form

4 0

3 years ago

Read 2 more answers

PLEASE HELP I REALLY NEED IT NOW. I WILL MARK AS BRAINLIEST

kotykmax [81]

Answer:

In Celcious, the temperature should be documented as C. 39.2 degrees.

Step-by-step explanation:

Here is the formula. I don't know if this will help but I'm trying.

Formula

(39.222°C × 9/5) + 32 = 102.6°F

;^)

6 0

3 years ago

HELP PLEASE ASAP <br><br> The decimal equivalent is a _____

sasho [114]

Terminating decimal

6 0

3 years ago

Read 2 more answers

Find $\int \dfrac{x}{\sqrt{1-x^4}}$ Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

6 0

3 years ago