All Answers 356090 Answers

Two ways to solve this.

The long way:

\((2+\frac{1}{2})^2\)

\((2+\frac{1}{2})(2+\frac{1}{2})\)

\(2\times2+2\times\frac{1}{2}+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\)

\(4+2\times\frac{1}{2}+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\)

\(4+\frac{2}{2}+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\)

\(4+1+2\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\)

\(4+1+\frac{2}{2}+\frac{1}{2}\times\frac{1}{2}\)

\(4+1+1+\frac{1}{2}\times\frac{1}{2}\)

\(4+1+1+\frac{1}{4}\)

\(5+1+\frac{1}{4}\)

\(6+\frac{1}{4}\)

\(\frac{24}{4}+\frac{1}{4}\)

\(\frac{25}{4}\)

\(6\frac{1}{4}\)

\(6.25\)

The short way:

\((2+\frac{1}{2})^2\)

\((\frac{4}{2}+\frac{1}{2})^2\)

\((\frac{5}{2})^2\)

\(\frac{5^2}{2^2}\)

\(\frac{25}{2^2}\)

\(\frac{25}{4}\)

\(6\frac{1}{4}\)

\(6.25\)

Press "2nd" key in the top left hand corner. You will see a key labled "asin", which is the same thing as inverse sine. Make sure you press = sign to get your answer.

what is 1,000,000,000 x 2,000,000,000=2 x 10^18

Because, you are probably doing something wrong. Enter 2 1/2 as 2.5 and press x^2 key followed by = sign to get your answer.

The Cosine MUST be between -1 and +1. You can't take the inverse cos of 9.26!!! 

$3,000 - $542 =$2,458 This is what Sarah needs in the next 4 months(net)

Now, will have to figure out the number of weeks in 4 months. Since there are 52 weeks in 1 year or 12 months, therefore:52/12 x 4 =17 1/3 weeks

$2,458.00 / (17 1/3) =$141.80 + $10 (for her expenses),

=$151.80 This is what Sarah needs to earn every week for about 4 months.

thanks

3/15  = 1/5  = 20%

81/49=10(10*49/81)

81/49 =10/ x      cross-multiply

81x = 49 *10

81x = 490       divide both sides by 81

x = 490/81

Split this up as two separate series

1 + 4 + 7 + 10 +....+ 301      and

-3  - 6  - 9  - 12 -.....- 300

The number of terms in the first series can be found as follows :

301 = 1 + (n - 1)3

300 = (n - 1)3

100 = n - 1

n = 101

So.....the sum of the first series = 

(101)(1 + 301)/2  =  15251

The number of terms in the second series can be found as follows:

-300 =  -3 - 3(n - 1)

-297  = -3(n - 1)

99  = n - 1

n = 100

So.......the sum of this series  =

(100)(-3 + -300)/2  =  - 15150

So......the sum of the two series =  15251  - 15150   =  101

Your final answer is 101, as Guest#1 calculated, NOT 303 that you say!.

1) Use this formula: FV=PV[1 + R]^N

FV=250[1 + 0.84/12]^5*12

FV=250 x 1.5197362

FV=$379.93

11) Use this formula:FV=P{[1 + R]^N - 1/ R}.[1+R]

FV=250{[1 + 0.084/12]^5*12 - 1/0.084/12}.[1+0.084/12]

FV=250  x [1.5197362 - 1 / 0.007] x 1.007

FV=350 x 74.248 x 1.007

FV=$18,691.94

As below?

.

I assume that this is ...... x ^4 + 7 by x - 3

            x^3   +  3x^2    +   9x        +  27

x - 3  [  x^4  +   0x^3    +   0x^2    +   0x    +   7   ].

            x^3  -    3x^3

          ________________________________

                        3x^3   +     0x^2

                        3x^2   -     9x^2

                       ______________

                                         9x^2      +   0x

                                         9x^2     -    27x

                                        ____________

                                                           27x    +   7

                                                           27x    -  81

                                                          _________

                                                                         88

So....the result is :      x^3   +  3x^2    +   9x    +  27   +   88 / [x - 3]

Try it this way:

.

this is a  quantitative analysis work
but i dont know how to use the formula to do

this is a math question

 It is noted that 8% of SP students are left handed. If 20 (TWENTY) students are randomly selected, calculate the

                 i. probability that none of them are left-handed, 

                 ii. probability that at most 2 are left-handed, 

                 iii. standard deviation for the number of left-handed students 

(b) If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN) classes have no left-handed students? 

(1)      tan(asin A)=((cosB)^2((tanB)^4-1)+1)/(tanB)^2

1-3+4-6+7-9+10-....-300+301

I will code it like

Look at the numbers with odd-numbered code which I have already bolded.

The numbers of code with the form 2n+1 are with the form of 3n+1.

Look at the numbers with even-numbered code which I set the font Italic.

The numbers of code with the form 2n are with the form 3n.

1-3+4-6+7-9+10-....-300+301

= (1+4+7+10+......+301)-(3+6+9+......+300)

There are \(\color{blue}\frac{301-1}{4-1}+1\) terms in (1+4+7+10+......+301)

which is 101 terms.

There are \(\color{blue}\frac{300-3}{6-3}+1\) terms in (3+6+9+......+300)

which is 100 terms.

1-3+4-6+7-9+10-....-300+301

= (1+4+7+10+......+301)-(3+6+9+......+300)

= \(\color{blue}\frac{(1+301)(101)}{2}-\frac{(3+300)(100)}{2}\)

=303 <------ Final Answer

\(tan(sin^{-1}A)=\frac{cos^2B(tan^4B-1)+1}{tan^2B}\)

Hi Max,

If you enter it in a form that I can just copy and past then I will enter it into Wolfram Alpha for you :)

(that means  no   LaTex)

It will show like

\(\frac{1}{3}+2=\frac{7}{3}=2.33333333333333333333\)

\({{10}^{10}}^{100}\times1000\)

= \({{10}^{10}}^{100}\times10^3\)

=\({10}^{({{10}^{100}+3)}}\)

The problem is: How many people are there in SP?

\(-\frac{2}{3}-(-2\frac{2}{5})\)

=\(-\frac{2}{3}+2\frac{2}{5}\)

=\(2\frac{2}{5}-\frac{2}{3}\)

=\(2\frac{6}{15}-\frac{10}{15}\)

= \(1\frac{21}{15}-\frac{10}{15}\)

=\(1\frac{11}{15}\)