All Answers 356032 Answers

For 2sin(x)dx, the 2 is just a constant, and since you know that dcos(x)/dx is -sin(x), then the integral is just -2cos(x).

\((x^{4y})^3\cdot2y^1\)

\(= x^{12y}\cdot 2y\)

\(=2x^{12y}y\)

\(\text{Also, thanks heureka and Alan for answering.}\)

\(\mbox{What if I got something like } du=2\sin x dx?\)

As below.  The trick is often in choosing the right terms for u and v.  I've used a different example from the one you suggested as it's fairly obvious that, in your example, the integral is just e^sin(x)  (as heureka has shown).

.

\(\begin{array}{rll}&&5\\-&&2\\&&3\end{array}\)

oh wait im already dumb .-.

XDDD

lol i feel so dumb now XDDD

5 - 2 = 3 .......

np :D

ty :)

\(\quad 15kg \div \dfrac{1}{6}\\ = 15kg \times 6 \\ = 90kg\)

His space suit and life support packs weigh 90kg on earth.

Actually, although I know how to do differentiation,
it's still a bit hard to get the original one from guessing by the
differentiated one.(I mean, doing integration)

How, actually, do I do integration by parts, for example,
I want to calculate

\(\int{(\cos x)(e^{\sin x})dx}?\)

\(\begin{array}{|lcll|} \hline \int{\cos x\cdot e^{\sin x}\ dx} \ ? \\ \text{Substitute: } u = e^{\sin x} \qquad du = e^{\sin x} \cdot \cos x \ dx \quad dx = \frac{du}{u\cdot \cos x} \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline && \int{\cos x\cdot e^{\sin x}\ dx} \\ &=& \int{\cos x\cdot u \cdot \frac{du}{u\cdot \cos x} } \\ &=& \int{\ du} \\ &=& u + c \\ \int{\cos x\cdot e^{\sin x}\ dx} &=& e^{\sin x} + c \\ \hline \end{array}\)

Oops it was googol. I was wrong. I didn't realize that. And 10 decillion was definitely 2.95 times less 0s that googol xD

It's 10 decillion.

\(a^2\equiv 0 \pmod{12}\\ \text{That means }\dfrac{a^2}{12}\mbox{ is a positive integer or 0}\\ \mbox{However }a\neq 0 \mbox{ so }\dfrac{a^2}{12} \mbox{ can't be 0.}\\ \begin{array}{rcl}\\&&a^2=12x\end{array}\\ \text{Check the multiples of 12 from 12 to 144}\\ \text{12 is not a square number}\\\text{24 is not a square number}\\\color{red}\text{36 is a square number} \\\text{48 is not a square number}\\ \)

\(\text{60 is not a square number}\\\text{72 is not a square number}\\\text{84 is not a square number}\\\text{96 is not a square number} \\\text{108 is not a square number}\\\text{120 is not a square number}\\\text{132 is not a square number}\\ \text{144 is not a square number}\)

\(\begin{array}{rllll}&a^2&=&36&\\&a&=&\pm 6&\\&\text{a must not be -6}\\\therefore&a=6.\end{array}\)

\(\huge{2)}\)

\(a^2\equiv 16 \pmod {10}\\a^2\equiv 6 \pmod {10}\\ a^2=\overline{X6}\;\;\mbox{<- This notation means 10X + 6 where }1\leq X\leq 9\\ \mbox{In 1 - 10 there are only 2 numbers whose squares end with 6, namely 4 and 6}\\ \therefore a=6\)

I guess you have spelt Christmas wrongly. And I think it's not Christmas today?

Other than the method Guest#1 suggested, you can just type atan(number) in the designated bar.

\(\text{Example:}\)

\( \begin{array}{rl}&\dfrac{\ln b^2}{2}\\=&\dfrac{2 \ln b}{2}\\=&\ln b\end{array}\)

I assume the case is , is the decimal point.

\(\begin{array}{rllll}\\4(38+x)&-&26,729&=&209,423\\&&4(38+x)&=&236,152 \\&&38+x&=&59,038\\&&x&=&21,038\end{array}\)

 cos^-1(1.55720185) is impossible as cosine function have a range of -1<x<1

4) \;\;Given\;\; a \equiv 1 \pmod{7},\;\;and\;\; b \equiv 2 \pmod{7} \'\'and\;\; c \equiv 6 \pmod{7} \mbox { what is the remainder when }  a^{81} b^{91} c^{27}   \mbox{is divided by 7}\?

\(4) \;\;Given\;a \equiv 1 \pmod{7},\;\;and\; b \equiv 2 \pmod{7} \:and\;\; c \equiv 6 \pmod{7}\\ \mbox { what is the remainder when }  a^{81} b^{91} c^{27}\;  \mbox{is divided by 7}?\)

Thanks   Heureka,

I have just finsihed Q4 too and I think our answers are fairly similar but I would like to post mine anyway :)

a=7p+1

b=7q+2

c=7r+6

p,q,r are integers

\(a^{81}=(7p+1)^{81}=1^{81}+81*(7p)^1+81C2*(7p)^2.........+(7p)^{81}\)

The only term here that is NOT a multiple of 7 is 1   and 1mod7=1

Using the same logic

\((7q+2)^{91}(mod 7) \\ =2^{91}(mod7)\\ =2^{10*9+1}(mod7) \\ =(2^{10})^{9}*2(mod7) \\ =(1024)^{9}*2(mod7) \\ =(2)^{9}*2(mod7) \\ =1*2(mod7) \\ =2(mod7) \)

and

\((7r+6)^{27}(mod7)\\ =6^{27}(mod7)\\ =((6^3)^3)^3(mod7)\\ =(216^3)^3(mod7)\\ =(6)^3(mod7)\\ =6(mod7)\)

so

\(a^{81} b^{91} c^{27} (mod7)\\ =1*2*6(mod7)\\ =12(mod(7)\\ =5 (mod7)\)

The remainder will be 5     Just like Heureka aready told us :)

Well... I got this to work haha.

And i also seem to have stumbled onto some form of an acceleration formula...

https://www.desmos.com/calculator/8m3s6uxrlb

Interesting!