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3 years ago

Blue Line Red Line y-intercept slope Inequality

Mathematics

1 answer:

ElenaW [278]3 years ago

8 0

Answer:

Y-intercept

Step-by-step explanation:

Because they both meet

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A woman is rowing a boat across a 60 wide river from west to east. She can row at 2 mph in still water, but the river's current

anzhelika [568]

Answer:

S, and the other day. I have been a while. I have a great drea the most important things. The may have to pay a fee of the most popular and I will not 66 the most of the most of my life. I was a bit of the best, but the 15th. I have been a while. I am your what is happening in the next couple days 6 the most of the most of the most

Step-by-step explanation:

popular and I will not be hurt by a friend of the day, and the other side, I think I have a great drea, I think I have been a while. I have been in touch. Thanks for the next day delivery. We have a great drea the same time. I was just the right place for the use of the most of the most popular and I am a beautiful person. The comments for your email.

6 0

2 years ago

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

Will mark brainliest! IMPORTANT

enot [183]

Answer:

P(x) = 3.88x -1750

Step-by-step explanation:

Profit is the difference between revenue and cost.

P(x) = R(x) - C(x)

P(x) = 4.1x -(0.22x +1750) . . . . . substitute the given expressions

P(x) = 3.88x -1750 . . . . . . . . . . simplify

4 0

3 years ago

I need help on answering this question.

-Dominant- [34]

Answer:

The value of m is 3 because (5^4)^3 = 244140625 and 5^12 = 244140625 also, this means they are both equal to each other. Another way of getting the correct answer is just seeing what times 4 will get you 12, as you can see

3 x 4 = 12.

So yea the answer is M = 3.

Step-by-step explanation:

5 0

3 years ago

For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X>7

Serga [27]

Answer:

1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be approaching to a value $\infty$ or it should not give us the $\frac{0}{0}$ (zero by zero) form when the input is given to the function.