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ZanzabumX [31]
2 years ago
14

Help me class8 (surds)​

Mathematics
1 answer:
Mashcka [7]2 years ago
3 0

Answer:

  • \cfrac{991\sqrt{2} }{80}

Step-by-step explanation:

For a start simplify each of the roots:

  • \sqrt{72} =\sqrt{36*2} =6\sqrt{2}
  • \sqrt{50} =\sqrt{25*2} =5\sqrt{2}
  • \sqrt{128} =\sqrt{64*2} =8\sqrt{2}
  • \sqrt{98} =\sqrt{49*2} =7\sqrt{2}

Now simplify the expression in steps:

\sqrt{72}-\cfrac{48}{\sqrt{50} }  -\cfrac{45}{\sqrt{128} }  +2\sqrt{98} =

6\sqrt{2}-\cfrac{48}{5\sqrt{2} }  -\cfrac{45}{8\sqrt{2} }  +2*7\sqrt{2} =

6\sqrt{2}-\cfrac{48*8+45*5}{5*8\sqrt{2} }  +14\sqrt{2} =

20\sqrt{2}-\cfrac{609}{40\sqrt{2} } =

20\sqrt{2}-\cfrac{609*\sqrt{2} }{40\sqrt{2} *\sqrt{2} } =

20\sqrt{2}-\cfrac{609\sqrt{2} }{40*2 } =

20\sqrt{2}-\cfrac{609\sqrt{2} }{80 } =

\sqrt{2} (20-7\cfrac{49}{80} )=

\sqrt{2} *12\cfrac{31}{80} =

\cfrac{991\sqrt{2} }{80}

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_________are triangles with congruent angles and proportional sides.
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Answer: Similar triangles

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Step-by-step explanation:

They have congruent angles and similar sides

7 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
Solve the following system using the SUBSTITUTION method.<br> x - 3y = 2<br> 2x - 6y = 6
notka56 [123]

ANSWER:

  • x – 3y = 2 ...(1)
  • 2x – 6y = 6 ...(2)

On solving eq(1), we get

  • x – 2 = 3y
  • x – 2/3 = y

Plug the value of x in eq(2)

  • 2x – 6(x - 2/3) = 6
  • 2x – 2(x – 2) = 6
  • 2x – 2x + 4 = 6
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So, given system of equations is not correct.

6 0
3 years ago
Read 2 more answers
15 points work out ratio for x
BlackZzzverrR [31]

Answer:

x = 25

Step-by-step explanation:

x : (x+10) = 5:7

Fractional form

x / x+10 = 5/7

Cross multiply:

x * 7 = (x+10) * 5

7x = 5x + 50

7x - 5x = 5x + 50 - 5x

2x = 50

x = 25

Check:

25 : 25 + 10

25 : 35

25/35 = 5/7

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

7 0
3 years ago
Can someone help me pls
algol [13]

Answer:

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Step-by-step explanation:

if you subtract each number from a number that is greater than all the numbers for example 200 then what ever has the most left will be greatest and least will be least for example 121 - 200 is negative 79 and 189 - 200 is negative 11 negative 11 is higher so its a greater number there is also easier ways to do this but this is just extra

8 0
2 years ago
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