Melody

Mmm 

I do not know the best way to do this but I will give it a go

13^1031 mod 599

13^1=13          = 13 mod 599

13^2=169        =  169 mod 599

13^3 = 2197     = 400 mod 599

13^4=28561    = 408 mod 588

13^5                = 512 mod 599

13^6                = 67 mod 599

13^7                = 272 mod 599

13^1031= 13^(6*171)*13^5

13^1031= ((67mod599)^171)*512mod599

13^1031= ((67mod599)^5)^34*(67mod599)*512mod599

13^1031= (72mod599)^34*(67*512mod599)

13^1031= (72mod599)^34*(161mod599)

13^1031= ((72mod599)^4)^8*(72mod599)^2*(161mod599)

13^1031= (320mod599)^8*(392mod599)*(161mod599)

13^1031= (320mod599)^3*(392mod599)^3*(320mod599)^2*(392mod599)*(161mod599)

13^1031= (304mod599)*(304mod599)*(570mod599)*(392*161mod599)

13^1031= ((304*304)mod599)   * (570mod599)*(217mod599)

13^1031= (170)mod599   *  (570*217)mod599

13^1031= (170)mod599   *  (296)mod599

13^1031= (170*296)mod599

13^1031   =     4   mod 599      

the answer is 4

Verification:

I assume there are 100 pennied in a dollar.

1200 pennies = 12.00 dollars

Hi Eight MersenePrime,

I was thinking about your username...   I looked up meridians to see where you might come from.

I am thinking that maybe you are from Strasbourg in France....  :)

I had not realized before that Bamako is further west than london - quite a lot further west !!

I love looking at maps :))

Now to answer your question :)

cos(19\pi/12)

\(cos(\frac{19\pi}{12})\\ \text{4th quadrant}\\ =+cos(\frac{(24-19)\pi}{12})\\ =+cos(\frac{5\pi}{12})\\ =+cos(\frac{3\pi}{12}+\frac{2\pi}{12})\\ =cos(\frac{\pi}{4}+\frac{\pi}{6})\\ =cos(\frac{\pi}{4})cos(\frac{\pi}{6})-sin(\frac{\pi}{4})sin(\frac{\pi}{6})\\ =cos(45^\circ)cos(30^\circ)-sin(45^\circ)sin(30^\circ)\\ =\frac{1}{\sqrt2}\times \frac{\sqrt3}{2}-\frac{1}{\sqrt2}\times \frac{1}{2}\\ =\frac{\sqrt3-1}{2\sqrt2}\\ =\frac{\sqrt3-1}{2\sqrt2}\times \frac{\sqrt2}{\sqrt2}\\ =\frac{\sqrt6-\sqrt2}{2*2}\\ =\frac{\sqrt6-\sqrt2}{4}\\ \)

How would you evaluate: (a^128 b^299 c^312)^2

                                            a^255 b^598 c^623

\(\frac{ (a^{128} b^{299} c^{312})^2}{a^{255} b^{598} c^{623}}\\~\\ =\frac{ a^{256} b^{598} c^{624}}{a^{255} b^{598} c^{623}}\\~\\ = a^{256-255} c^{624-623}\\~\\ =ac\)

The total for a baseball glove is $105.93, which includes a 7% sales tax. What is the original price of the glove?

Original Price = P

P+7% of P = 105.93

1P+0.07P = 105.93

1.07P=105.93

P=105.93/1.07

105.93/1.07 = $99

I do not understand your working but I got the same answer:

It makes the equation TRUE .

Ich verstehe nicht Ihre Arbeit, aber ich bekam die gleiche Antwort:
Es macht die Gleichung WAHR

\(3log (5x) + 7 log (3x) = 15 \\ log (5x)^3 + log (3x) ^7= 15 \\ log ((5x)^3 * (3x) ^7)= 15 \\ log (125x^3 * 2187x^7)= 15 \\ log (273375x^{10})= 15 \\ log (273375)+logx^{10}= 15 \\ logx^{10}= 15-log (273375) \\ 10logx= 15-log (273375) \\ logx= {15\over10}-{log (273375) \over10}\\~\\ x=10^\left( {15-log (273375) \over10}\right)\\ x=9.043\)

check

x=9.0432413241285286124

Does       3 log (5*9.0432413241285286124) + 7 * log (3*9.0432413241285286124) = 15 

3*log(5*9.0432413241285286124)+7*log (3*9.0432413241285286124) = 14.9999999999999992

That is 15

this makes the equation true :)

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Maybe you could explain to me better how you did it???  Anyway, your answer is correct :)

Vielleicht könntest du mir besser erklären, wie du das gemacht hast ??? Wie auch immer, Ihre Antwort ist richtig :)

That is correct EP  :)

1)

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

int of x^2sinx dx

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

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.