What system of equations represents the graph?

Mathematics

1 answer:

il63 [147K]3 years ago

4 0

B. Because of the slope and where it the lines intersect the y axis

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Choose the set that consists of only prime numbers.

butalik [34]

The answer is C because 2, 5, 23, and 29 are prime numbers (you can also look up a chart and highlighted numbers are prime). prime numbers is a whole number greater than 1 and cannot be formed when multiplying two smaller numbers that are whole. it must involve itself in a multiplication equation.

6 0

3 years ago

The movies ticket cost $24 per person. Anna and her sister both go to the movies. There is a long line to get into the theatre.

Mumz [18]

Answer:

multiply the amount of people by 24 so far all we know is that there are 2 people in line so for rn it would be $48

Step-by-step explanation:

4 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

Simplify the complex fraction. 2 1/2 / 5/8

lbvjy [14]

First do 2 1/2 / 5/8
5/2/ 5/8=2/5 * 5/8=1/4
now you can do anything with 1/4 so
your answer is 1/4

5 0

3 years ago

Read 2 more answers

Solve: x^+4/x-1 = 5/x-1

AlexFokin [52]

The first step for solving this equation is to determine the defined range.
$\frac{ x^{4} }{x-1} = \frac{5}{x-1}$ , x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:
$x^{4}$ = 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/- $\sqrt[4]{5}$
Separate the solutions.
x = $\sqrt[4]{5}$ , x ≠ 1
x = - $\sqrt[4]{5}$
Check if the solution is in the defined range.
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
This means that the final solution to your question are the following:
x = $\sqrt[4]{5}$
x = - $\sqrt[4]{5}$
Let me know if you have any further questions.
:)

7 0

3 years ago