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9/11

multiplication is basically adding up a certain number a certain amount of times. 2 times 2 is basically 2+2.

Don't get it? Here's an example.

If you have 2 friends who both have 2 cookies, how many cookies are there in total? 2+2=4.

The same works with other numbers, for example, 4 times 3, which is just adding 4 3 times, or 4+4+4=12

Another real life example.

There are 4 fruits in a basket, how many fruits are in 3 baskets?

4+4+4 =12  

Thanks!

The last choice is the correct one......

Because 10 is 10^1, 10^6 is just adding 6 zeros after 1.

3.3 * 1000000 = 3300000

(Just move the decimal point 6 digits to the right and fill the empty spots with zero.

Correct, NSS !!!!!

I'm assuming that you have

 (5+2.5sqrt(3))/(25/4*sqrt(3) )   =

5 / (25/4*sqrt(3) )    +  2.5sqrt (3) / (25/4 * sqrt (3) )  =

5 / (6.25 sqrt (3) )   +  2.5sqrt (3) / ( 6.25 * sqrt (3) )   =

5 / (6.25 sqrt (3) )   +  2.5 / ( 6.25  ) 

.8 / sqrt (3)   +  .4      =        rationalize the denominator in the frist fraction

.8sqrt (3) / [ sqrt (3) * sqrt (3) ]  +   .4  =

.8 sqrt (3) / 3  +   .4   =              get a common denominator

[.8sqrt (3)  + .4 (3)]  / 3  =

 [ .8 sqrt (3)  + 1.2 ]  / 3  =

.4 [ 2sqrt (3)  + 3 ]  / 3  =

(.4/3) [ 2sqrt (3)  + 3 ]  =

(4/30) [ 2sqrt (3) + 3 ] =

(2/15) [ 2sqrt (3) + 3  ]  ≈  .861880

Writing the vector components thusly :

A  = < 2cos 0° , 2sin 0° >

B  =  < 3.75 cos 90°, 3.75 sin 90° >

C  =  < 2.5 cos 180°. 2.5 sin 180° >

D = < 3 cos 270°, 3 sin 270° >

Sum the x components  = 2cos 0° + 3.75cos 90° + 2.5cos 180° + 3cos 270°  = 2 + 0 - 2.5 + 0  =  -.5

Sum the y components  = 2sin 0° + 3.75sin 90° +  2.5 sin 180° + 3 sin 270° = 0 + 3.75 + 0 - 3  = .75

Resultant vector  =  < x , y >  =  < -.5, .75 >     →  in the second quadrant

Resultant magnitude  = sqrt ( x^2 + y^2 )  =  sqrt [ (-.5)^2  + (.75)^2 ] ≈  . 90134 km

Direction  =   tan^-1 (y/x)  = tan^-1 ( .75 / -.5)  = tan^-1 ( -3/2) = tan^-1 (-1.5)  ≈ -56.31°

However....this is a second quadrant angle so, with respect to due west, it is 56.31° north of due west 

[i.e.  123.69°  in standard position  ]

The first "set" of three zeros is the hundreds.

Also, you can call them sets of zeros as I don't believe there is a name for them. 

(6+n):5=-10

1.2+0.2n=-10

0.2n=-11.2

n=-56

13. Really?

$(3.5)(4.9)(1.7) = 29.155$

13, did you really not know?

and g 

What is the number for f

It is impossible

Can someone answer this: (1-3x)*(2+i), and also explain the FOIL system?

I do not usually teach this method but I shall explain it since you ask.

(1-3x)*(2+i)

The first bracket has 2 terms, they are                  1 and -3x

The second bracket also has 2 terms, they are     2 and   i

F stands for first that is 1*2 = 2

O stands for outside so that is   1*+i = i

I stands for inner                      -3x*2 = -6x

L stands for last                       -3x*i = -3xi

Add them all together and you have

$(1-3x)(2+i)\\=2+i+-6x+-3xi\\ =2+i-6x-3xi$

i is often used to denote an imaginary number but since this is a low level question I assume that i is a pronumeral the same as x is.  :)

An infinite geometric series has first term -3/2 and sums to twice the common ratio.

Find the sum of all possible values for the common ratio.

If the common ratio r lies between −1 to 1 , we can have the sum of an infinite geometric series.

That is, the sum exits for $| r |<1$ .

The sum S of an infinite geometric series with \( -1 is given by the formula, $S=\frac{a_1}{1-r}$

Let $a_1 = -\frac{3}{2}$

Let $r =\,?$

$\begin{array}{|rcll|} \hline S = \frac{a_1}{1-r} &=& 2r \quad & | \quad a_1 = -\frac{3}{2} \\ \frac{-\frac{3}{2}}{1-r} &=& 2r \\ -\frac{3}{2} \cdot \frac{1}{1-r} &=& 2r \quad & | \quad : 2 \\ -\frac{3}{4} \cdot \frac{1}{1-r} &=& r \\ \frac{3}{4} \cdot \frac{1}{r-1} &=& r \quad & | \quad \cdot (r-1) \\ \frac{3}{4} &=& r \cdot (r-1) \quad & | \quad - \frac{3}{4} \\ r \cdot (r-1) - \frac{3}{4} &=& 0 \\ r^2-r- \frac{3}{4} &=& 0 \\\\ r &=& \frac{1\pm\sqrt{1-4\cdot (- \frac{3}{4}) } }{2} \\ r &=& \frac{1\pm\sqrt{1+3 } }{2} \\ r &=& \frac{1\pm 2 }{2} \\\\ r_1 &=& \frac{1 + 2 }{2} \\ r_1 &=& \frac{3}{2} \quad & | \quad \text{ no common ratio, because } r > 1 \\\\ r_2 &=& \frac{1 - 2 }{2} \\ r_2 &=& -\frac{1}{2} \\ \hline \end{array}$

the sum of all possible values for the common ratio is $-\frac12$

By definition the angle in radians is given by arc length/radius, i.e. $\theta = s/r$

So, for 01:  $\theta = 6/27 \rightarrow 0.222 \text{ radians}$

See if you can now do the others (don't forget to convert degrees to radians first in 03 and 04).

What is the value of the sum 1/1*3 + 1/3*5 + 1/5*7+ 1/7*9+...+1/199*201?

Express your answer as a fraction in simplest form.

$\begin{array}{rcll} && \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7}+ \frac{1}{7*9}+\ldots+\frac{1}{199*201} \\ &=& \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7}+ \frac{1}{7*9}+\ldots+\frac{1}{(2n-1)(2n+1)} \\ \hline && \frac{1}{(2n-1)(2n+1)} = \frac12\left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \\ && \frac{1}{1*3} = \frac12\left( \frac{1}{1} - \frac{1}{3} \right) \\ && \frac{1}{3*5} = \frac12\left( \frac{1}{3} - \frac{1}{5} \right) \\ && \frac{1}{5*7} = \frac12\left( \frac{1}{5} - \frac{1}{7} \right) \\ && \frac{1}{7*9} = \frac12\left( \frac{1}{7} - \frac{1}{9} \right) \\ && \ldots \\ && \frac{1}{199*201} = \frac12\left( \frac{1}{199} - \frac{1}{201} \right) \\ \hline &=& \frac12\left( \frac{1}{1} - \frac{1}{3} \right) + \frac12\left( \frac{1}{3} - \frac{1}{5} \right) + \frac12\left( \frac{1}{5} - \frac{1}{7} \right) + \frac12\left( \frac{1}{7} - \frac{1}{9} \right)+\ldots+\frac12\left( \frac{1}{199} - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{1}{1} - \underbrace{\frac{1}{3} + \frac{1}{3}}_{=0} - \underbrace{\frac{1}{5}+\frac{1}{5}}_{=0} - \underbrace{\frac{1}{7} + \frac{1}{7}}_{=0} - \underbrace{\frac{1}{9}+ \frac{1}{9}}_{=0} +\ldots- \underbrace{\frac{1}{199}+\frac{1}{199}}_{=0} - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{1}{1} - \frac{1}{201} \right) \\ &=& \frac12\left( 1 - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{201-1}{201} \right) \\ &=& \frac12\left( \frac{200}{201} \right) \\ &=& \frac{100}{201} \\ \end{array}$

Compute 1*1/2+2*1/4+3*1/8+...+n*1/2n+...

$\begin{array}{|rcll|} \hline S_n &=& 1*\frac{1}{2}+2*\frac{1}{4}+3*\frac{1}{8}+4*\frac{1}{16}+\ldots+n*\left(\frac{1}{2}\right)^n+\ldots \\ \hline \end{array}$

(i)

$\text{Multiply } S_n \text{ by } \tfrac12 \text{ the common ratio of the geometric series}$

$\small{ \begin{array}{rcccccccccccccl} S_n &=& 1*\frac{1}{2} &+& 2*\frac{1}{4} &+& 3*\frac{1}{8} &+& 4*\frac{1}{16} &+& \ldots &+& n*\left(\frac{1}{2}\right)^n && \\ \frac12 S_n &=& & & 1*\frac{1}{4} &+& 2*\frac{1}{8} &+& 3*\frac{1}{16} &+& \ldots &+& (n-1)*\left(\frac{1}{2}\right)^n &+& n*\left(\frac{1}{2}\right)^{n+1} \\ \hline (1-\frac12)S_n &=& [~ 1*\frac{1}{2} &+& 1*\frac{1}{4} &+& 1*\frac{1}{8} &+& 1*\frac{1}{16} &+& \ldots &+& 1*\left(\frac{1}{2}\right)^n ~] &-& n*\left(\frac{1}{2}\right)^{n+1} \quad (\text{subtract})\\ \end{array} }$

The series in the square brackets is a geometric series with $a = \frac12$, $r = \frac12$ and $n$ terms,
Thus, $S_n$ for this series =$\dfrac{a(1-r^{n})}{1-r} = \dfrac{\frac12(1-(\frac12)^n)}{1-\frac12}=1-(\frac12)^n$

$\begin{array}{rcll} (1-\frac12)S_n &=& [~ 1*\frac{1}{2} + 1*\frac{1}{4} + 1*\frac{1}{8} + 1*\frac{1}{16} + \ldots + 1*\left(\frac{1}{2}\right)^n ~] - n*\left(\frac{1}{2}\right)^{n+1} \\ (1-\frac12)S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ \end{array}$

(ii)

$\text{Because } \left|\frac12\right| < 1 \text{, then } \lim \limits_{n\to \infty} { \left(\frac12 \right)^n } = 0 \text{ and } \lim \limits_{n\to \infty} { \left(\frac12 \right)^{n+1} } = 0$

$\begin{array}{rcll} \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ S_n &=& 2 \left( 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \right) \\ \lim \limits_{n\to \infty} {S_n} &=& 2 \left( 1-0 - n*0 \right) \\ &=& 2\cdot( 1 ) \\ &=& 2 \\ \end{array}$

-6(3p-1) = (-6*3p)+(-6*-1) = -18p+6

If I did not misunderstand then,

1

In the first example the sum of the numbers is 1. There also only is one number in the first example.

Therefore in the first example the sum of all numbers is equal to the amount of numbers. 

Huh?  What kind of an answer is that?  If you do not know the answer to these questions, then do not answer.