All Answers 356020 Answers

4+416/12=?

4+34 2/3=38 2/3

2^nd

This is a trig question.

And there's one information missing. What's the angle between the hoop and the ground?

jake wants to use 250 grams of butter. how many grams of flour and how many grams of sugar will he need?

recip: 406 gr flour, 276 gr butter, 113 gr sugar.

If we use proportionality, then we have:

250/276 x 406 =~367.75 gr. of flour.

250/276 x 113=~102.35 gr. of sugar.

jake wants to use 250 grams of butter. how many grams of flour and how many grams of sugar will he need?

recip: 406 gr flour, 276 gr butter, 113 gr sugar.

(155°)/(360°)×π×2×7.3 ≈19.75

what is 7x10 to the -24 power

(7x10)^-24 =5.219717 x 10^-45

3x10 to the power of -10

30^(-10) =1.69350 x 10^-15

n+(n+1)+(n+2)+(n+3) =162

4n+6 =162

4n=162 -6

4n =156

n=156/4

n=39, therefore:

39 + 40 + 41 + 42 =162

n + 2+n + 4+n =120

3n + 6 =120

3n =120 - 6

3n=114

n=114/3

n=38

38+2=40

38+4=42

38 +40 +42 =120

Find a closed form for

S_n=1*1!+2*2!+...+n*n!.

for integers n>=1.

Your response should have a factorial.

Example for \(n=4\):

\(\begin{array}{|rcll|} \hline S_4 &=& 1*1!+2*2!+3*3!+4*4! \quad & | \quad 4! = 3!*4 \\ S_4 &=& 1*1!+2*2!+3*3!+4*3!*4 \\ S_4 &=& 1*1!+2*2!+3!*(3+4*4) \\ S_4 &=& 1*1!+2*2!+3!*(3+4^2) \quad & | \quad 3! = 2!*3 \\ S_4 &=& 1*1!+2*2!+ 2!*3*(3+4^2) \\ S_4 &=& 1*1!+2!*[~2+ 3*(3+4^2)~] \quad & | \quad 2! = 1!*2 \\ S_4 &=& 1*1!+1!*2*[~2+ 3*(3+4^2)~] \\ S_4 &=& 1!*\{~1+2*[~2+ 3*(3+4^2)~]~\}\quad & | \quad 1! = 1 \\ S_4 &=& 1*\{~1+2*[~2+ 3*( \underbrace{3+4^2}_{ \underbrace{=4-1+4^2}_{ \underbrace{=4*(4+1)-1}_{=4*5-1} } })~]~\} \\ S_4 &=& 1*\{~1+2*[~2+ 3*( 4*5-1 )~]~\} \\ S_4 &=& 1*[~1+2*(~2+ 3*4*5-3 ~)~] \\ S_4 &=& 1*[~1+2*(~3*4*5-1 ~)~] \\ S_4 &=& 1*(~1+2*3*4*5-2 ~) \\ S_4 &=& 1*(~ 2*3*4*5-1 ~) \\ S_4 &=& 1*2*3*4*5-1\\ S_4 &=& 5!-1\\ 1*1!+2*2!+3*3!+4*4! &=& 5!-1\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline S_n &=& 1*1!+2*2!+...+n*n! \\ S_n &=& (n+1)!-1 \\ \hline \end{array}\)

I’m missing the link that frames the dialogue’s perspective. 

Here’s the missing link: (We chimps are exceptional at finding missing links)

\(h'(1)=\dfrac{1}{\sqrt{1+1^2}}=\dfrac{\sqrt2}{2}\)

\(g'(1)=(e^1)((1+1^2)(1+1)+2(1)^2)=6e\)

\(f'(1)= 1^2-15(1)^2(\ln 1)= 1\)

\(\frac{d}{dx}\sqrt{1+x^2}\\\begin{array}{rl}h=\sqrt{u}&u=1+x^2\\\end{array}\\ \frac{dh}{dx}=\frac{dh}{du}\times \frac{du}{dx}=\dfrac{1}{2\sqrt{1+x^2}}\times 2x=\dfrac{x}{\sqrt{1+x^2}}\)

\(\frac{d}{dx} (1 + xe^x (1+x^2))\)

\(= \frac{d}{dx} 1 + (1+x^2)\frac{d}{dx}xe^x+(xe^x)\frac{d}{dx}(1+x^2)\)

\(= (1+x^2)(e^x)(1+x)+(x)(e^x)(2x)\)

\(=(e^x)((1+x^2)(1+x)+2x^2)\)

\(\frac{d}{dx}(1-x^3(5\ln x + 2))\)

\(=\frac{d}{dx}1 - \frac{d}{dx} 5x^3\ln x +\frac{d}{dx} 2x^3\)

\(=\frac{d}{dx}2x^3-(5x^3)\frac{d}{dx}\ln x - (\ln x)\frac{d}{dx} 5x^3 \)

\(= 6x^2 - (5x^3)(\frac{1}{x})-(\ln x )(15x^2)\)

\(= 6x^2-5x^2-15x^2\ln x\)

\(= x^2-15x^2\ln x\)

There is a \(\pm\) button under the '2' button. That is your +/-

\(\dfrac{16-48}{-2}\\ =\dfrac{48-16}{2}\\ =\dfrac{32}{2}\\ =16\)

\(\dfrac{log_3\frac{3}{4}+log_3\frac{4}{5}+log_3\frac{5}{6}+log_3\frac{6}{7}+log_3\frac{7}{8}+log_3\frac{8}{9}}{log\frac{4}{5}+log\frac{5}{6}+log\frac{6}{7 }+log\frac{7}{8}+log\frac{8}{9}+log\frac{9}{10}}\)

=\(\dfrac{log_33-log_34+log_34-log_35+log_35-log_36+log_36-log_37+log_37-log_38+log_38-log_39}{log4-log5+log5-log6+log6-log7+log7-log8+log8-log9+log9-log10}\)= \(\dfrac{log_33-log_39}{log4-log10}\)

=\(\dfrac{1-2}{\log\dfrac{2}{5}}\)

=\(-\log\dfrac{2}{5}\)

Same as CPhill's answer.

\(10^2\div 0.0541\\ =100 \div \dfrac{541}{10000}\\ = 100 \times \dfrac{10000}{541}\\ =\dfrac{1000000}{541}\\ \approx 1828.429\)

http://web2.0calc.com/questions/i-couldn-t-find-my-question-so#r1

I am using this too :D I love this website that I can teach people to do Maths and being taught to do maths :D