Melody

Hi JM,

I am stumped to at the moment.

Are you sure there isn't more information like maybe PSRQ is a parallelogram?

1)  a+b+c+d+e+f+g+h+i+j+k+l = 24.05

2)  (a+b+c+d+e+f) - (g+h+i+j+k+l) = 6.05

3)  9(a+b+c+d+e+f) / (g+h+i+j+k+l) = 15.05

4)  (a+b+c+g+h+j) - (d+e+f+i+j+l) = 2.95

5)  (a+c+e+g+i+k) - (b+d+f+h+i+j) = 1.95

6)  2(d+e+f+i+j+l) - (a+b+c+g+h+j) = 7.6

7)  abc + def + ghi + jkl = 40.978

8)  (abc + ghi) - (def - jkl) = 19.642

9)  ab + cd + ef + gh + ij + kl = 28.48

10)  (ab + ef + ij) - (cd + gh + kl) = 12.52

11)  (a+b)*(c+d) + (e+f)*(g+h) + (i+j)*(k+l) = 48.45

12)  (a+b)*(c+d) - (e+f)*(g+h) - (i+j)*(k+l) = 8.55

a=? b=? c=? d=? e=? f=? g=? h=? i=? j=? k=? l=?

Let     x = a+b+c+d+e+f           and   y = g+h+i+j+k+l

x+y=24.05   From (1)

x-y=6.05      From (2)

add

2x=30.1

x=15.05

y=24.05-15.05 = 9

so

lets look at (3)

9x/y = 9*15.05/9 = 15.05     so (3) is true and it does not help solve anything.

So we have exhausted all the clues from  lines 1,2 and 3 and we have discovered that:

a+b+c+d+e+f = 15.05      and          g+h+i+j+k+l = 9

My interest has now wained, maybe I will come back to it later.    :)

Of course you can post it Juriemagic  :D

\(|x-5|<-1 \)

If you are talking real numbers that is impossible.

The absolute value of anything has to be greater or equal to 0.

It can not be less than a negative number!

Thanks Adam,

Here NinjaDevo gave a wonderful answer for 3-6

How many distinct, non-equilateral triangles with a perimeter of 60 units have integer side lengths $a$, $b$, and $c$ such that $a$, $b$, $c$ is an arithmetic sequence?

let the sides by a, a+d, a+2d    where a and d are both positive integers and d>0

\(a+a+d+a+2d=60\\ 3a+3d=60\\ a+d=20\\ \)

9 of the combinations work :)

What good manners :)

If the distance between (3, -2) and (4, a)  is the sqaure root of 7, find the exact value of a.

\(\sqrt{(4-3)^2+(a--2)^2}=\sqrt7\\ (4-3)^2+(a--2)^2=7\\ 1+(a+2)^2=7\\ (a+2)^2=6\\ a+2=\pm\sqrt6\\ a=-2\pm\sqrt6\\ \)

Hi Guest, 

I really like that you have explained what you did :)

Lets take a look.

Question:

How mnay 3-digit numbers can be constructed from the digiits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used:

a. As often as desired: 343 

Lets do units*tens* hundreds

7*7*7

b. Only once: 210  

There is no zero so any digit can be anywhere.

7*6*5

I couldn't do C, I'd assume that you divide 210 by 2 to get 105. But the answer is 120, what am I missing out here?

c. Once only and the number must be odd?

The units this time can be 1 or 3 or 5 or 7  That is 4 possibilities

So that is  4*6*5 = 120

does that help?

Thanks Heureka,

Guest, it would be polite to first say thank you to Heureka.

And then to explain why your think one of his answers is incorrect!

Maybe, no answer here comes with a guarentee. You have to learn, from us and others, so that you can pick up any careless errors for yourself  :)