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a camera cost 115 Canadian dollars. If each Canadian dollar is worth 0.952 U.S. dollars, how much will the camera cost in US dollars? please help me with this problem is it dividing addition subtraction or multiplying

\(\begin{array}{|rcll|} \hline && 115 ~\text{Canadian dollars} \times \frac{0.952 ~\text{U.S. dollars} }{1 ~\text{Canadian dollar} } \\ &=& 115 \times 0.952 ~\text{U.S. dollars} \\ &=& 109.48 ~\text{U.S. dollars} \\ \hline \end{array} \)

0.0256 = e^(.46t) how to solve this equation ?? steps please

\(\begin{array}{|rcll|} \hline 0.0256 &=& e^{(0.46\cdot t)} \qquad & | \qquad ln() \\ \ln(0.0256) &=& 0.46\cdot t \qquad & | \qquad : 0.46 \\ \frac{ln(0.0256)}{0.46 } &=& t \\ t &=& \frac{ln(0.0256)}{0.46 } \\ t &=& \frac{-3.66516292750}{0.46 } \\ t &=& -7.96774549456 \\ \hline \end{array} \)

Greetings GingerAle,

Thank you very much.

Viele Grüße Heureka

Greetings GingerAle,

Thank you very much.

Viele Grüße Heureka

Greetings GingerAle,

Thank you very much.

Viele Grüße Heureka

30C5

\(\begin{array}{|rcll|} \hline && 30C5 \\ &=& \binom{30}{5} \\\\ &=& \dfrac{30!}{(30-5)!\cdot 5! } \\\\ &=& \dfrac{30!}{ 25! \cdot 5! } \\\\ &=& \dfrac{25!\cdot 26 \cdot27 \cdot 28 \cdot 29 \cdot 30}{25!\cdot 5! } \\\\ &=& \dfrac{ 26 \cdot27 \cdot 28 \cdot 29 \cdot 30}{ 5! } \\\\ &=& \dfrac{30}{5} \cdot \dfrac{29}{4} \cdot \dfrac{28}{3} \cdot \dfrac{27}{2} \cdot \dfrac{26}{1} \\\\ &=& \dfrac{30}{5} \cdot \dfrac{29}{1} \cdot \dfrac{28}{4} \cdot \dfrac{27}{3} \cdot \dfrac{26}{2} \\\\ &=& 6 \cdot 29 \cdot 7 \cdot 9 \cdot 13 \\\\ &=& \mathbf{142506} \\ \hline \end{array} \)

Is there a pattern to 0,6,24,60,120,210? This is fiendishly hard. Or how about 0,4,10,20,35. Please save me and solve!

\(\begin{array}{|r|rcr|} \hline n & n\cdot(n+1)\cdot(n+2) \\ \hline 0 & 0\cdot 1 \cdot 2 &=& 0 \\ 1 & 1\cdot 2 \cdot 3 &=& 6 \\ 2 & 2\cdot 3 \cdot 4 &=& 24 \\ 3 & 3\cdot 4 \cdot 5 &=& 60 \\ 4 & 4\cdot 5 \cdot 6 &=& 120 \\ 5 & 5\cdot 6 \cdot 7 &=& 210 \\ 6 & 6\cdot 7 \cdot 8 &=& 336 \\ 7 & 7\cdot 8 \cdot 9 &=& 504 \\ 8 & 8\cdot 9 \cdot 10 &=& 720 \\ \cdots & \cdots \\ \hline \end{array}\)

!1+1

\(\begin{array}{|rcll|} \hline !1 &=& 1! \cdot \left( 1- \frac{1}{1!} \right) \\ &=& 1 \cdot \left( 1- 1 \right) \\ &=& 1 \cdot 0 \\ &=& 0 \\\\ !1+1 &=& 0 + 1 \\ &=& 1 \\ \hline \end{array}\)

ABC is a right angle triangle. D is the point on AB such that AD=3DB. AC=2DB and angle A=90.

Show that sinC=k/square root 20, where k is an interger. Write down the value of k.

\(\begin{array}{|rcll|} \hline DB &=& x \\ AD &=& 3x \\ AB &=& AD + DB\\ &=& 3x +x \\ &=& 4x\\\\ AC &=& 2DB \\ &=& 2x \\\\ hypotenuse &=& CB \\ CB^2 &=& AB^2+AC^2 \\ &=& (4x)^2+(2x)^2\\ &=& 16x^2+4x^2\\ &=& 20x^2 \\ CB &=& \sqrt{20x^2} \\ &=& x\cdot \sqrt{20}\\\\ \sin(C) &=& \frac {AB}{CB} \\ &=& \frac {4x}{x\cdot \sqrt{20}} \\ &=& \frac {\mathbf{4} } { \sqrt{20} } \\ \text{So the value of k is }\mathbf{4} \\ \hline \end{array} \)

determine the sum of all the multiples of 4 between 1 and 999

arithmetical sequence: \(sum = 4 + 8 + 12 + 16 + 20 + 24 + \dots+996\)

\(n = \frac{996}{4} = 249\)

\(\begin{array}{|rcll|} \hline sum &=& \frac{(a_1+a_n)\cdot n}{2} \\ &=& \frac{(4+996)\cdot 249}{2} \\ &=& \frac{1000\cdot 249}{2} \\ &=& 500 \cdot 249 \\ &=& 124500 \\ \hline \end{array}\)