If u answer i gives u my remaing poaints

If f(x) equals 3X -2 what is f(5.9)

Ivanshal [37]

Answer:

f(5.9)= 15.7

Step-by-step explanation:

f(x) = 3x -2

f(5.9) = 3(5.9) -2

f(5.9) = 17.7 -2

f(5.9) = 15.7

Word Explination

If the (x) in f(x) is replaced by another variable (in this case, a number) any other x seen in the equation is replaced by that variable. For this question, the x in f(x) is replaced by 5.9. And so any other x in this equation needs to be replaced by 5.9 as well. From there on out, you solve.

6 0

3 years ago

What is the greatest whole number that rounds to 2800 when rounded to the nearest hundred? The least whole number? What is the

guapka [62]

add me on fortnite IClapYsn

8 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

Alexis cell phone company charges her $35 a month for phone service plus $0.40 for each text message. How many text message does

r-ruslan [8.4K]

Answer:

$92.50

Step-by-step explanation:

$72 - $35 = 37

37 / 0.40 = 92.5

Answer = $92.50

8 0

3 years ago

In the first quadrant you start at 6,8 and move 3 units right 2 units down

Finger [1]

Answer:

You would end up at (9,6)

8 0

3 years ago