All Answers 356032 Answers

1)

\(\int\;\frac{x}{x^2+6x+10}\;dx\\ =\frac{1}{2}\int\;\frac{2x+6}{x^2+6x+10}\;dx-\int\;\frac{-3}{x^2+6x+10}\;dx\\ =\frac{ln|x^2+6x+10|}{2}+3\int\;\frac{1}{x^2+6x+9+1}\;dx\\ =\frac{ln|x^2+6x+10|}{2}+3\int\;\frac{1}{(x+3)^2+1}\;dx\\ =\frac{ln|x^2+6x+10|}{2}+3tan^{-1}(x+3)+c\\\)

Thanks a lot!

Thanks a lot!

Thanks a lot!

int of x^2sinx dx

\(\int x^2sinx\;dx\\ \qquad \text{Integrate by parts}\\ \qquad u=x^2\qquad v'=sinx\\ \qquad u'=2x\qquad v=-cosx\\ =[-x^2cosx]-\int -2xcosx\;dx\\ =[-x^2cosx]+2\int xcosx\;dx\\ \qquad \text{Integrate by parts}\\ \qquad u=x\qquad v'=cosx\\ \qquad u'=1\qquad v=sinx\\ =[-x^2cosx]+2\{(xsinx)-\int sinx\;dx\}\\ =[-x^2cosx]+2\{xsinx+cosx\;\}+c\\ =-x^2cosx+2xsinx+2cosx+c\\ =(2-x^2)cosx+2xsinx+c\)

1)

Take the integral:
 integral x/(x^2 + 6 x + 10) dx
Rewrite the integrand x/(x^2 + 6 x + 10) as (2 x + 6)/(2 (x^2 + 6 x + 10)) - 3/(x^2 + 6 x + 10):
 = integral((2 x + 6)/(2 (x^2 + 6 x + 10)) - 3/(x^2 + 6 x + 10)) dx
Integrate the sum term by term and factor out constants:
 = 1/2 integral(2 x + 6)/(x^2 + 6 x + 10) dx - 3 integral1/(x^2 + 6 x + 10) dx
For the integrand (2 x + 6)/(x^2 + 6 x + 10), substitute u = x^2 + 6 x + 10 and du = (2 x + 6) dx:
 = 1/2 integral1/u du - 3 integral1/(x^2 + 6 x + 10) dx
The integral of 1/u is log(u):
 = (log(u))/2 - 3 integral1/(x^2 + 6 x + 10) dx
For the integrand 1/(x^2 + 6 x + 10), complete the square:
 = (log(u))/2 - 3 integral1/((x + 3)^2 + 1) dx
For the integrand 1/((x + 3)^2 + 1), substitute s = x + 3 and ds = dx:
 = (log(u))/2 - 3 integral1/(s^2 + 1) ds
The integral of 1/(s^2 + 1) is tan^(-1)(s):
 = (log(u))/2 - 3 tan^(-1)(s) + constant
Substitute back for s = x + 3:
 = (log(u))/2 - 3 tan^(-1)(x + 3) + constant
Substitute back for u = x^2 + 6 x + 10:
Answer: |= 1/2 log(x^2 + 6 x + 10) - 3 tan^(-1)(x + 3) + constant

2)

Take the integral:
 integral x^2 sin(x) dx
For the integrand x^2 sin(x), integrate by parts, integral f dg = f g - integral g df, where
 f = x^2, dg = sin(x) dx, df = 2 x dx, g = -cos(x):
 = -x^2 cos(x) + 2 integral x cos(x) dx
For the integrand x cos(x), integrate by parts, integral f dg = f g - integral g df, where
 f = x, dg = cos(x) dx, df = dx, g = sin(x):
 = -x^2 cos(x) + 2 x sin(x) - 2 integral sin(x) dx
The integral of sin(x) is -cos(x):
 = x^2 (-cos(x)) + 2 x sin(x) + 2 cos(x) + constant
Which is equal to:
Answer: |= 2 x sin(x) - (x^2 - 2) cos(x) + constant

2. \int x^2sinxdx ?

Formula:

\(\begin{array}{|rcll|} \hline \int(u\cdot v')\ dx = u\cdot v - \int(u'\cdot v)\ dx \\ \hline \end{array}\)

\(\begin{array}{|rclrclrcl|} \hline && \int (~x^2\cdot \sin(x) ~)\ dx\\ && & u &=& x^2 & u' &=& 2x \\ && & v' &=& \sin(x) & v &=& -\cos(x) \\ &=& -x^2\cdot \cos(x) + 2\cdot\int (x\cdot \cos(x) )\ dx \\ \\ \hline && \int (~x\cdot \cos(x) ~)\ dx\\ && & u &=& x & u' &=& 1 \\ & & & v' &=& \cos(x) & v &=& \sin(x) \\ &=& x\cdot \sin(x) - \int (1\cdot \sin(x) )\ dx \\ &=& x\cdot \sin(x) - \int (\sin(x) )\ dx & \int \sin(x) &=& -\cos(x) \\ &=& x\cdot \sin(x) + \cos(x) \\ \\ \hline && \mathbf{ \int (~x^2\cdot \sin(x) ~)\ dx }\\ &=& -x^2\cdot \cos(x) + 2\cdot[x\cdot \sin(x) + \cos(x) ] + c \\ &=& -x^2\cdot \cos(x) + 2\cdot x\cdot \sin(x)+ 2\cos(x)+ c \\ &=& \mathbf{ 2\cdot x\cdot \sin(x) -(x^2-2)\cdot \cos(x) + c } \\ \hline \end{array}\)

Thank you very much, CPhill !

What a strange question!

You don't have to...no one can make you!

You can drop out of school.

You can truant.

you can pay no attention.

I am quite sure you hve a heap of other options as well !!

If you do not want to learn maths, you should consider not attending a maths forum...

Just a thought.     

What is the smallest number that 

when divided by 31 leaves a remainder of 9, and

when divided by 41 leaves a remainder of 39?

\(\begin{array}{|rcll|} \hline n \equiv {\color{red}9} \pmod {{\color{green}31}} \\ n \equiv {\color{red}39} \pmod {{\color{green}41}} \\ m=31\cdot 41=1271 \\ \hline \end{array} \)

Because 31 and 41 are relatively prim ( gcd(31,41) = 1! ) we can go on:

\(\begin{array}{rcll} n &=& {\color{red}9} \cdot {\color{green}41} \cdot [ \frac{1}{\color{green}41}\pmod {{\color{green}31}} ] + {\color{red}39} \cdot {\color{green}31} \cdot [ \frac{1}{\color{green}31}\pmod {{\color{green}41}} ] + {\color{green}31}\cdot {\color{green}41} \cdot k \quad & | \quad k\in Z\\ \end{array}\)

\(\begin{array}{rcll} && [ \frac{1}{\color{green}41}\pmod {{\color{green}31}} ] \quad &| \quad \text { modular inverse } 41\cdot \frac{1}{41} \equiv 1 \pmod {31} \\ &=& {\color{green}41}^{\varphi({\color{green}31})-1} \pmod {{\color{green}31}} \quad & | \quad \text{ Euler's totient function } \varphi(n) \quad \varphi(p) = p-1 \\ &=& 41^{30-1} \pmod {31} \\ &=& 41^{29} \pmod {31} \quad & | \quad 41 \pmod {31} = 10 \\ &=& 10^{29} \pmod {31} \\ &=& 28 \\\\ && [ \frac{1}{\color{green}31}\pmod {{\color{green}41}} ] \quad &| \quad \text { modular inverse } 31\cdot \frac{1}{31} \equiv 1 \pmod {41} \\ &=& {\color{green}31}^{\varphi({\color{green}41})-1} \pmod {{\color{green}41}} \quad & | \quad \text{ Euler's totient function } \varphi(n)\quad \varphi(p) = p-1 \\ &=& 31^{40-1} \pmod {41} \\ &=& 31^{39} \pmod {41} \\ &=& 4 \\ \end{array} \)

\(\begin{array}{|rcll|} \hline n &=& {\color{red}9} \cdot {\color{green}41} \cdot 28 + {\color{red}39} \cdot {\color{green}31} \cdot 4 + {\color{green}31}\cdot {\color{green}41} \cdot k \\ n &=& 10332 + 4836 + 1271 \cdot k \\ n &=& 15168 + 1271 \cdot k \quad & | \quad k\in Z \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline n_{min} &=& 15168 \pmod {1271 } \\ n_{min} &=& 1187 \\\\ n &=& 1187 + 1271 \cdot k \quad & | \quad k\in Z \\ \hline \end{array}\)

The smallest number is 1187

What is the last digit of: 3^2046?

\(\begin{array}{|rcll|} \hline && 3^{\varphi(10) } \equiv 1 \pmod{10} \quad | \quad \varphi(10) = 10\cdot(1-\frac12)\cdot(1-\frac15) = 4 \\ && 3^4 \equiv 1 \pmod{10} \\\\ && 3^{2046} \pmod{10} \\ &\equiv & 3^{4\cdot 511 +2} \pmod{10} \\ &\equiv & 3^{4\cdot 511}\cdot 3^2 \pmod{10} \\ &\equiv & (3^4)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & (1)^{511}\cdot 3^2 \pmod{10} \\ &\equiv & 3^2 \pmod{10} \\ &\equiv & 9 \pmod{10} \\ &\equiv & 9 \\ \hline \end{array} \)

The last digit is 9

What is the smallest number under 16,000 that

when divided by 127 leaves a remainder of 14,

and when divided by 131 leaves a ramainder of 66.

\(\begin{array}{|rcll|} \hline n \equiv {\color{red}14} \pmod {{\color{green}127}} \\ n \equiv {\color{red}66} \pmod {{\color{green}131}} \\ m=127\cdot 131=16637 \\ \hline \end{array} \)

Because 127 and 131 are relatively prim ( gcd(127,131) = 1! ) we can go on:

\(\begin{array}{rcll} n &=& {\color{red}14} \cdot {\color{green}131} \cdot [ \frac{1}{\color{green}131}\pmod {{\color{green}127}} ] + {\color{red}66} \cdot {\color{green}127} \cdot [ \frac{1}{\color{green}127}\pmod {{\color{green}131}} ] + {\color{green}127}\cdot {\color{green}131} \cdot k \quad & | \quad k\in Z\\ \end{array} \)

\(\begin{array}{rcll} && [ \frac{1}{\color{green}131}\pmod {{\color{green}127}} ] \quad &| \quad \text { modular inverse } 131\cdot \frac{1}{131} \equiv 1 \pmod {127} \\ &=& {\color{green}131}^{\varphi({\color{green}127})-1} \pmod {{\color{green}127}} \quad & | \quad \text{ Euler's totient function } \varphi(n) \quad \varphi(p) = p-1 \\ &=& 131^{126-1} \mod {127} \\ &=& 131^{125} \mod {127} \\ &=& 32 \\\\ && [ \frac{1}{\color{green}127}\pmod {{\color{green}131}} ] \quad &| \quad \text { modular inverse } 127\cdot \frac{1}{127} \equiv 1 \pmod {131} \\ &=& {\color{green}127}^{\varphi({\color{green}131})-1} \pmod {{\color{green}131}} \quad & | \quad \text{ Euler's totient function } \varphi(n)\quad \varphi(p) = p-1 \\ &=& 127^{130-1} \mod {131} \\ &=& 127^{129} \mod {131} \\ &=& 98 \\ \end{array}\)

\(\begin{array}{|rcll|} \hline n &=& {\color{red}14} \cdot {\color{green}131} \cdot 32 + {\color{red}66} \cdot {\color{green}127} \cdot 98 + {\color{green}127}\cdot {\color{green}131} \cdot k \\ n &=& 58688 + 821436 + 16637 \cdot k \\ n &=& 880124 + 16637 \cdot k \quad & | \quad k\in Z \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline n_{min} &=& 880124 \pmod {16637 } \\ n_{min} &=& 15000 \\\\ n &=& 15000 + 16637 \cdot k \quad & | \quad k\in Z \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline k=0: & \mathbf{n} &\mathbf{=}& \mathbf{15000} \\ \hline \end{array} \)

The smallest number is 15000

Sorry Chris, I was having a brain collapse   

\(42_{10}=4 \;\;mod 19\\ 24_{19}=42_{10}\\\)

I often get mod and base muddled up but I usually correct it before I go public.   

I like the trig approach, heureka......!!!!!

Let ABCD be a convex quadrilateral,

and let P, Q, R, S, T, U, V, and W be the trisection points of the sides of ABCD, as shown. 

If the area of quadrilateral ABCD is 180, then find the area of hexagon AQRCUV.

Let AB = a
Let BC = b
Let CD = c
Let DA = d

Let \(\angle{ABC}\) = B
Let \(\angle{UDV}\) = D

\(\begin{array}{|rcll|} \hline Area_{\text{ABC}} &=& \frac{ab\sin(B)}{2} \\ Area_{\text{CDA}} &=& \frac{cd\sin(D)}{2} \\ Area_{\text{ABCD}} &=& Area_{\text{ABC}} + Area_{\text{CDA}} \\ &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline Area_{\text{QBR}} &=& \frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} \\ Area_{\text{UDV}} &=& \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} - Area_{\text{QBR}} - Area_{\text{UDV}} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\left(1-\frac19 \right) + \frac{cd\sin(D)}{2}\left(1-\frac19 \right) \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\cdot \frac89 + \frac{cd\sin(D)}{2}\cdot \frac89 \\ Area_{\text{hexagon}} &=& \frac89 \cdot \left( \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \right) \\ Area_{\text{hexagon}} &=& \frac89 \cdot Area_{\text{ABCD}} \\ Area_{\text{hexagon}} &=& \frac89 \cdot 180 \\ \mathbf{ Area_{\text{hexagon}} } & \mathbf{=} & \mathbf{160} \\ \hline \end{array}\)

The area of hexagon AQRCUV is 160

18!/4!(18-4)!=

\(\begin{array}{|rcll|} \hline && \frac{18!}{4!(18-4)!} \\ &=& \binom{18}{4} \\ &=& \frac{18}{4}\cdot \frac{17}{3}\cdot \frac{16}{2}\cdot \frac{15}{1} \\ &=& \frac{18}{2}\cdot \frac{17}{1}\cdot \frac{16}{4}\cdot \frac{15}{3} \\ &=& 9\cdot 17\cdot 4\cdot 5 \\ &=& \mathbf{3060} \\ \hline \end{array} \)

if i have the numbers -6 5 4 2 and have to use bedmas and use all of the numbers once to get to -10 ?

\(\begin{array}{|rcll|} \hline -10 &=& [5-6-4] \cdot 2 \\ -10 &=& \frac{-6-4}{2}-5 \\ \hline \end{array}\)

and a seperate question same instructions -22 how do i do it?

\(\begin{array}{|rcll|} \hline -22 &=& (5-2)\cdot (-6)-4 \\ -22 &=& \frac{(-6-5)\cdot4}{2} \\ \hline \end{array}\)

2(19)^1 + 4(19)^0  =

38 + 4  =

42  ....!!!!

(6cos 30º)²

= \([6(\frac{ \sqrt{3}}{2})]^{2}\)

= \((3\sqrt{3})^{2}\)

= (9)(3)

= 27

Mmmm

42 base 19 = 4   not 24

there is no base that works Chris.

Maybe guest was using alian maths.   ://

1 whole and      1         is what?

                        ---- %

                         4

\(1\frac{1}{4}\%=1.25\%=0.0125\)

You graph your equations and the solution is where they cross.

Here is a beautiful graphing calculator.

just use brackets

example

6^(2/3)

Wassup!

One plus one equals two. If you can't read that, 1+1=2