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4. In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other?

[ I learned  this trick from Melody......]

Let us consider the number of arrangements where Wilma  and Paul are sitting together

We can seat them in any of 6 positions and arrange the other people in 5! ways......but.....we can also arrange Wilma and Paul in two different ways in each of the 6 positions.....so.......the total arrangements where they sit together = 6 * 2 * 5!  =  1440 ways

And the total number of ways to seat all 7 people = 7!  = 5040 ways

So.......the total number of ways to seat all 7 people  minus the arrangements where Wilma and Paul sit togeher  = the number of arrangements where they don't sit together

5040 - 1440  =  3600 arrangements where they are not sitting together.......

4                                                                          

3. I have 6 shirts, 6 pairs of pants, and 6 hats. Each item comes in the same 6 colors ( so that I have 1 item of each color). I refuse to wear an outfit in which all 3 items are the same color. How many choices for outfits do I have?

Assuming that two of the items can be the same color.......

We have  6*6*6   = 216 possible outfits if we can wear any color that we'd like........however....only 6 of these are all the same color......

So.....the total number of outfits you could wear, but avoiding all the same colors  =  216 - 6  = 210

Each hat matches up with 6 shirts, so ((1 hat)*(6 shirts)*(6 hats)) gives 36 shirt/hat combos. Each shirt/hat combo matches with 5 pants, so in total, there are 180 combos. There have to be 5 of one so that none have all one color. Or, 180=6!/4

There are  99,999 - 10,000 + 1  = 90,000 five digit numbers

Let's just consider how many of them have no zeroes.......

We have 9 ways to select a non-zero digit for each of the positions, so this must be  9^5  = 59049  five digit numbers that have no zeroes........then  90,000 - 59,049   = 30,951 of them must contain at least one zero

1. How many 4-letter words with at least one vowel can be constructed from the letters A,B,C,D, and E?

Assuming that we can have "non-sense" words

We could have  a word with an "A"  and this letter could appear in any of 4 positions......and we could choose any of the other three consonants to occupy a second positon, then any two of the consonants to occupy a third position, and the last consonant would appear in the last position by default.......so we have

4 * 3 * 2 *1  =      24 words using just the A  and the other three consonants

The same result would occur if we just used the "E" and the other three consonants

If we used the "A"  and the "E," we would have 24 words using any two of the other consonants......but, we could choose any 2 of the 3 remaining consonants.....so   the total "words"   would be  24 * (3C2)  = 24 * 3  =  72

So....the total number of words would  be 24 + 24 + 72  =  120 "words"

Hi,

the rendered formula font has a fixed size, same for all users. Implementing a configuration  for this would not be easy.

Does Melody's  suggestion work for you?

Best regards,

Admin

There is an easier way to solve the problem. You know that the triangle is a right triangle because it's sides are a Pythagorean triple. Saying this we can call the point perpendicular to B as X. Then we will call AX = y and CX = 26 - y. BX we will call a. Then using the Pythagorean Theorem y ^ 2 + x ^ 2 = 10^2, y^2 + x^2 = 100. Then we can make another equation using the other triangle. (26 - y)^2 + x ^ 2 = 24^2. This simplifies to x^2 + y^2 - 52y = -100. Plugging in information from the other equation 100 - 52y = -100. y = 50/13. Then plugging this in x = 120/13

sorry i posted the same answer twice. so sorry

pi is a number that basically goes on 4ever. but here is what your calculator might show you if you have a scientific one. you wil see a symbol that looks like this but but the squiggly is above the two lines ~|| if you press that you should get something like the following as the answer: 3.1415926535897932 most people just use 3.14 as pi. hoope this helps! 

Since x = 5 is a solution, replace  x  with  5  in the equation:  x² - x - m  =  0

     --->     (5)² - (5) - m  =  0     --->     25 - 5 - m  =  0     --->    20 - m  =  0     --->     -m  =  -20     --->     m  =  20

If you put your cursor over the (log) key, you can access the (ln) key.

You must use the (ln) key in your calculation, not the letters 'l' and 'n':

Enter as:  ln(2)/ln(1.00416667)/12

I am sure I speak for MaxWong and our answering guests as well as for myself when I say that we are very pleased that we are helping you.

Welcome to the Web2.0 forum Kcc2468  

We are here to help people learn kcc.

It is best that you are up front about what kind of help that you need.

We don't have time to answer everyone's questions so help us to know which of yours have priority.

There is no need for us to redo all your homework unless you have real concerns that you have not really understood.

So tell us what you did and let us comment on your work.

You will get better reactions from us long term if we feel you are using our expertise and time to learn the from us in the best, most efficient manner :))

We knock ourselves out for students who show us very clearly that they are learning from us and not using us for anything other than learning purposes.  :)

I just  copied it straight into the web2.0calc

[(2*-3)^2*6-50]/2 = 83

(x-5)(3x+25) 

x(3x+25)     +    -5(3x+25)

Now expand :)    remember that the minus sign ALWAYS belongs to the 5

Recentl;y asked - already answered.

if you have an equation like 6x+12= 8x-24 what would that equal

It will only be equal if

6x+12= 8x-24

12= 2x-24

36= 2x

x=18

It will not be equal for any other value of x  :)

I have them done I'm just looking for solutions to ensure I'm right! Thank you melody for help! 

Only Mr Massow can answer your question

BUT

can't you just magnify your screen ?

Max: Everything appears to be right in your calculations except this one:

(c)  Find the final value of a savings account into which e 1000 are paid every three months for 10 years, if the bank is paying 2.5% p.a. interest compounded quar- terly.

FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD.

FV=1,000[1 + .025/4]^(10*4) -1/(.025/4)}

FV=1,000{[1.283027 - 1] / .00625}

FV=1,000 x 45.284291

FV=45,284.29e

YES and these are all square numbers too :))

Good thinking guest :)

Do you understand it up to here  ?

\((x-2)(x^2+bx+c)=x^3-4x^2-11x+30\)

The bit in the box is just equating coefficients.

On the LHS the x^2 terms are      \(bx^2\;\; and \;\;-2x^2\)  

and on the right the x^2 term is      \( -4x^2\)

so

\(bx^2\;\;\ +\;\; -2x^2\;=\;-4x^2\\ (b-2)x^2=-4x^2\\ b-2=-4\\ b=-2\\ \)

Is that all you need help with ?

Thank you melody and guest!