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oksian1 [2.3K]
3 years ago
11

243.875 to ne rest tenth, hundredth, ten and hundred

Mathematics
1 answer:
shutvik [7]3 years ago
6 0
243.9 tenth
243.88 for hundredth
240 for ten
200 for hundred
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1. 2 (2x +1) + x = x + 2x + 18<br> 2. 2x + x + 2x = 10 + x
GuDViN [60]

Answer:

1: X= 8, 2: 2/5 or 0.4

Step-by-step explanation:

1: 2(2x + 1) + x = x + 2x + 18

5x + 2 = 3x + 18

2x = 16

x = 8

2: 2x + x + 2x = 10 + x

5x = 10 + x

4x = 10

x = 4/10

x = 2/5

x = 0.4

8 0
2 years ago
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Consider the function that assigns to each positive number x the value
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2 years ago
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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}&#10;\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\&#10;&&\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)}\\\\\\&#10;\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}\\\\\\&#10;\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\&#10;\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)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\&#10;=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\&#10;=\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
HELP PLS ASAP I HAVE 6 MINUTES NO LINKS
Ainat [17]


you could say if you knew how many cubes there were in total, you could subtract them
5 0
2 years ago
Answer please hurry i need it
Snezhnost [94]

Answer:

0

Step-by-step explanation:

x+12 = 9+2x +3

x = 12-12

x = 0

4 0
2 years ago
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