Simplify the following: -i^2 i^2 = -1: --1 (-1)^2 = 1: Answer: | | 1

If the first number in a number line is x^2, and the second number is x^2+1, and the third number is X^2+2, then CALCULATE the sum in terms of "x".

\(\begin{array}{lrcll} \text{sum of } & ~ x^2 ,x^2 +1,x^2+2 ~\text{ is } ~3x^2 + 3 \\ \text{factored i asume} & ~ 3(x^2 + 1) \end{array}\)

You keep missing the key work in your (original)  question.

SUM

Such a shame..... He seemed to rather be looking forward to it.....

x*3/4=72*1/3 

x*3/4=72/3

x=72*4/3*3

x=288/9

x=32

There are som comments that are just plain stupid... That was an irrelevant comment... You couldn't have just simply thanked Huereeka?

Noooooo..really????..so easy?...gosh, thanx Heureka...I bet sometimes you feel your level of intelligence is mocked????...

Round the next number to 1) Second decimal and 2), Third decimal. The number is \(2.456715 * 10^{12}\).

\(\begin{array}{lrcll} 1) \quad \text{ Second decimal } & 2.46 * 10^{12}\\ 2) \quad \text{ Third decimal } & 2.457 * 10^{12}\\ \end{array}\)

3/4 of a number is 72.what is 1/3 of the number?

\(\begin{array}{rcll} x \cdot \dfrac34 &=& 72 \qquad & | \qquad \cdot \frac43 \\ x &=& 72 \cdot \frac43 \\ x &=& 96 \\ x &=& 96 \qquad & | \qquad \cdot \frac13 \\ x\cdot \frac13 &=& 96 \cdot \frac13 \\ \mathbf{ x\cdot \frac13 } &\mathbf{=}& \mathbf{ 32 }\\ \end{array} \)

1/3 of the number is 32

Coldplay hath drawn my attention to this post.

Sorry to dissappoint Mistress Coldpay, as Head Mistress my time was much called upon.

The flea bitten cur will have to wait at least till the sun doth rotate about the Earth once again before he can be the H Mistress's valantine.  

In the mean time he could woo his chosen with flowers and chocolate and pre chewed bones. 

Just a thought Nauseated :)

Thanx Melody, I do appreciate...

If the first number in a number line is x^2, and the second number is x^2+1, and the third number is X^2+2, then CALCULATE the sum in terms of "x".

a=x^2

d=1

\(S_n=\frac{n}{2}[2a+(n-1)d]\\ S_n=\frac{n}{2}[2x^2+(n-1)1]\\ S_n=\frac{n}{2}[2x^2+n-1]\\\)

I don't know what they mean about factoring - maybe it is already factored.

THIS IS A PRESENT VALUE FORMULA

4/4+4/5+4/6+4/7+7/8

= (4*210 + 4*168 + 4*140 + 4*120 + 7*105 ) / (2³ * 3 * 5 * 7)

=  3287 / 840

asinus :- ) !

thank you Nauseated !

Did you send any special Valentine's cards melody? I believe nauseated had requested one... Was his wish granted? Hahahahaha

Indeed? Perhaps I did get swept away in the excitement... Or maybe I simply ignored your council, for lack of better words... I shall go find some Valentine's Day posts...:)

Please give at least the next 5 terms of this sequence...........

4, 7, 8, 11, 16, 19, 24, 27, 38, 41.........etc.

I asume:

\(\begin{array}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline &4&{\color{green}7}&8&11&16&19&24&2{\color{green}7}&38&41&{\color{grey}48}&{\color{grey}53}&{\color{grey}56}&{\color{grey}61}&{\color{grey}68} \\ \hline \text{Prime numbers}&2&3&{\color{blue}5}&{\color{blue}7}&{\color{blue}11}&{\color{blue}13}&{\color{blue}17}&{\color{blue}19}&23&29&31&37&41&43&47 \\ \text{Prime numbers}&{\color{red}3}&{\color{red}5}&7&11&13&17&19&23&{\color{red}29}&{\color{red}31}&{\color{red}37}&{\color{red}41}&{\color{red}43}&{\color{red}47}&{\color{red}53}\\ \hline \text{consecutive numbering}&{\color{red}1}&{\color{red}2}&{\color{blue}3}&{\color{blue}4}&{\color{blue}5}&{\color{blue}6}&{\color{blue}7}&{\color{blue}8}&{\color{red}9}&{\color{red}10}&{\color{red}11}&{\color{red}12}&{\color{red}13}&{\color{red}14}&{\color{red}15}\\ \hline \text{add red or add blue }& \\ \text{change with digit seven} &\\ \hline \end{array}\)

What is the probability of  8 successes in 15 trials, assuming independent trials with probability of success p=.6. Thanks for the help.

15C8* (0.6^8)*(0.4^7)

Integrate: dx/sqrt[1 - x^2], for all x from -1 to 1.

\(\small{ \begin{array}{lrcll} &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } \\ \hline \text{Substitution : } & x &=& \sin{(u)} \qquad \rightarrow \qquad u = \arcsin{(x)}\\ & \ dx &=& \cos{(u)}\ du \\ \hline &\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-\sin^2{(u)}} } \ dx } \qquad | \qquad 1-\sin^2{(u)} = \cos^2{(u)}\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \sqrt{\cos^2{(u)}} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} } \ dx } \qquad | \qquad dx = \cos{(u)}\ du\\ & &=&\int \limits_{-1}^{1} { \frac{1}{ \cos{(u)} }\cdot \cos{(u)}\ du } \\ & &=&\int \limits_{-1}^{1} { \ du } \\ & &=& [ u ]_{-1}^{1} \qquad | \qquad u = \arcsin{(x)} \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } & \mathbf{=} & \mathbf{ [~ \arcsin{(x)} ~]_{-1}^{1} }\\ \hline & \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } & = & [~ \arcsin{(x)} ~]_{-1}^{1} \\ & & = & [~ \arcsin{(1)}-\arcsin{(-1)} ~]\\ & & = & [~ \frac{\pi}{2} -\frac{-\pi}{2} ~]\\ & & = & [~ \frac{\pi}{2} +\frac{\pi}{2} ~]\\ & & = & \pi \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ \pi } \\ & \mathbf{ \int \limits_{-1}^{1} { \frac{1}{ \sqrt{1-x^2} } \ dx } } &\mathbf{=}& \mathbf{ 3.14159265359 } \end{array} }\)

Solve for k:

3 (4 k+14) = 10 k-2 (k-7)

12k + 42 =10k - 2k + 14

12k + 42 =8k + 14

12k - 8k =14 - 42

4k = -28 divide both sides by 4

k = -7

Look at the graph and note that strain C has the least virulence in March. Use this point in time and calculate the weighted average to determine the efficacy of the vaccine.

To calculate the weighted average, take the (March) prevalence of each strain and multiply it by the efficacy of the vaccine in the table. Then add these products. 

\((0.35 * 0.23) + (0.13 * 0.25) + (0.76 * 0.13) + (0.68 * 0.39) =0.447 ~|~(47.7\%)\)

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Now if only we could develope a vaccine for CDD. 

Divide both sides by -2

7.5 miles = 

7.5 * 5280 * 12   = 475,200  inches....so...

475,200/25  = 19008 strides