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3 years ago

Y=3x−5 and y=x−1 Is (2,1) a solution of the system?

Mathematics

1 answer:

zlopas [31]3 years ago

7 0

Answer:

Yes (2,1) is a solution to the equation

Step-by-step explanation:

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A boy spent one-half of his money for a book and one-third of his money for a pen. The remaining $2.25 he saved. How much money

Tema [17]

Let his original money is x

So, $\frac{x}{2}$ + $\frac{x}{3}$ + 2.25 = x
or, $\frac{3x+2x+(6*2.25)}{6} = x$
or, 5x + 13.5 = 6x
or, x = 13.5

So, he had $13.5 originaly.

Thanks!

8 0

3 years ago

Read 2 more answers

The graph only goes up to 10 FYI

Otrada [13]

Answer:

This is the first line.

point one (0,-2)

point two (5,-5)

This is the second line.

point one (0,6)

point two (1,7)

3 0

3 years ago

A taxi cab in myrtle beach charges $2 per mile and $1 for every person. If a taxi cab ride for two people costs $12, how far did

Oxana [17]

Answer:

5 Miles

Step-by-step explanation:

5 Miles = 1 Miles = $2 X 5 = $10

1 Person = $1

Therefore 2 People = $1 X 2

6 0

3 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

Brandon is 6 times as old as Cora. In 4 years, Brandon will be only twice as old as

nalin [4]

$C-Cora's\ age\ now\\B-Brandon's\ age\ now\\\\B=6C\ \ \ and\ \ \ B+4=2(C+4)\\\\6C+4=2C+8\\6C-2C=8-4\\4C=4\ \ \ \Rightarrow\ \ \ C=1\ \ \ \Rightarrow\ \ \ \ B=6\cdot1=6\\\\Ans.\ Brandon\ is\ now\ 6\ years.$

4 0

3 years ago

Y=3x−5 and y=x−1 ​ Is (2,1) a solution of the system?

Y=3x−5 and y=x−1 Is (2,1) a solution of the system?