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Correctamundo! 

(xi – x)2 where xi = 1, 2, 3, 4, 5

Mmm

Do you mean

\(\displaystyle\sum_{\xi=1}^5\;(\xi-5)^2\quad?\)

\(\displaystyle\sum_{\xi=1}^5\;(\xi-5)^2\\ =(1-5)^2+(2-5)^2+(3-5)^2+(4-5)^2+(5-5)^2\\ =16+9+4+1+0\\ =30\)

Thanks XSquared   

Chris has already answered and that is great but I have a suggestion.

How about people answer on a new question thread and then put in an answer link.

That way the answer is not right with the question. It is more like having to look up the back of the book for the answer.

This might not work for a lot of people and that is fine, it is just a suggestion   

4√5 (√ [3j ] + 8 )  - 3√ [ 15j ]  + √5

4 √ [ 15j ] + 32√5 - 3√ [15j] + √5

√ [ 15j ] + 33√5

The leg of a right triangle is 1/3  the sum of the other sides, and the perimeter is 12 units. What is the area of this traingle?

Let a, b  be the legs and c the hypotenuse

So

a^2 + b^2  =  c^2   ⇒     a^2 = c^2  - b^2 ⇒   a  = sqrt ( c^2 - b^2 )

3a  = b + c

a + [b + c]  = 12

a + 3a  =  12

4a  = 12

a  = 3

This implies that    c^2 - b^2  =  9

And b + c = 9 

So

c^2 - b^2 =  b + c

(c - b) (c + b) =   (b + c)     divide through by  (b + c)

So 

(c - b)  =  1  ⇒   c = b + 1

So

b + ( b + 1)  = 9

2b = 8

b = 4        and c  = 5

And the area is just the  product of the legs / 2   =   a * b / 2 =  3 * 4 / 2  =  6 units^2

2y + 3x = 8

1y - 2x = -3

Multiply the second equation through by -2 

2y + 3x  = 8

-2y + 4x  = 6      add these

       7x  =  14     diviede both sides by 7

         x  = 2

And using 1y - 2x = -3  to find y

1y  - 2(2)  = -3

y  - 4   = -3         add 4 to both sides

y =  1

\(0^0=0^{1-1}=0^1*0^{-1}=\frac{0}{0}\)

In other words, \(0^0\) is really a disguised case of \(\frac{0}{0}\), which is also undefined. Therefore, 0^0 does not have an assigned value to it.

Here's one to begin the thread! 

The leg of a right triangle is \(\frac{1}{3}\) the sum of the other sides, and the perimeter is 12 units. What is the area of this traingle?

Good luck to any participants!

Hi Jeff

You maths is fine but you have not made either x or y the subject.

Yours is fine, you only had to put x on the left hand side  

what is (y-5)/2<-x/3

Make x the subject:

\((y-5)/2<-x/3\\ \frac{3(y-5)}{2}<-x\\ \frac{-3(y-5)}{2}>x\\ x<\frac{-3(y-5)}{2}\\ or\\ x<\frac{3(5-y)}{2}\\\)

Make y the subject

\(\frac{y-5}{2}<\frac{-x}{3}\\ y-5<\frac{-2x}{3}\\ y<-\frac{2x}{3}+5\\ \)

3x - 5y = -5

2x + 3y = 22

Multiply the frist equation through by 3   and the second through by 5  ..... and we have that

9x   -  15y  =  - 15

10x + 15y =    110      add these

19x  = 95      divide both sides by 19

x  =  5

And using 2x + 3y = 22 to find y, we have

2(5) + 3y  = 22

10 + 3y  = 22      subtract 10 from both sides

3y  = 12              divide both sides by 3

y  = 4

\(\frac{-4}{-24}\,=\,\frac{-4\div-4}{-24\div-4}\,=\,\frac16\)

Thanks, AdminMod....I knew it looked familiar....LOL!!!!!

0^0 is undefined.

Nice, hectictar.....!!!!

You have to make this more clear. Are we solving x in terms of y or vice versa? I am going to assume that it is in terms of y.

What is \(\frac{y-5}{2}<-\frac{x}{3}\) ?

Multiply 6 on both sides to get \(3y-15<-2x\)

Divide -2 on both sides, but we have to change the sign \(-\frac{3}{2}y+\frac{15}{2}>x\)

Prime numbers are numbers that can only be divided by 1 and itself. The first five prime numbers are 2, 3, 5, 7, 11, and so on.

Composite numbers are numbers that are divisible by integers other than 1 and itself. The first five composite numbers are 4, 6, 8, 9, 12

1 is not considered a prime number or a composite number.

1/12

volume of sphere  \(=\,\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3\)

volume of hemisphere  \(=\,\frac12\,*\,\frac43\,*\,\pi\,*\, (\text{radius of hemisphere})^3\)

Plug in  \(3\sqrt[3]2\)  for the radius of the hemisphere.

volume of hemisphere  \(=\,\frac12\,*\,\frac43\,*\,\pi \,*\,(3\sqrt[3]2)^3 \\~\\ =\,\frac23\,*\,\pi\,*\,3^3\,*\,\sqrt[3]2^3 \\~\\ =\,\frac23\,*\,\pi\,*\,27\,*\,2 \\~\\ =\,36\pi\)

The hemisphere has the same volume as the sphere, so....

                         volume of sphere  =  volume of hemisphere

\(\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3\,=\,36\pi\)

                                                                         Divide both sides by  pi .

\(\frac43\,*\, (\text{radius of sphere})^3\,=\,36\)

                                                                         Multiply both sides by  3/4  .

\((\text{radius of sphere})^3\,=\,27\)

                                                                         Take the cube root of both sides.

\(\text{radius of sphere}\,=\,3\,\text{ cm}\)

The way you have your formula written, t is for half-life, which is the exact opposite of doubling time!

Solve for t over the real numbers:
1200/e^(0.025 t) = 600

1200/e^(0.025 t) = 1200 e^(-t/40):
1200/e^(t/40) = 600

Divide both sides by 1200:
e^(-t/40) = 1/2

Take reciprocals of both sides:
e^(t/40) = 2

Take the natural logarithm of both sides:
t/40 = log(2)

Multiply both sides by 40: 
t = 40 log(2) =~ 27.726 Time units.

????? wut

https://web2.0calc.com/questions/a-circle-is-drawn-that-intersects-all-three-sides-of-triangle-pqr-as-shown-below-prove-that-if-ab-cd-ef-then-the-center-of-the-circle-is-the

nvm

i didnt see the 1st 0

its 1

The given function transforms the parent function, y = abs (x),  in the following ways

a. Reflects the function across the x axis [ due to the negative sign]

b. Vertically "compresses" the function by (1/2)

c. Shifts this end result up by 3 units

"Vertex"  =  ( 0,3)

Domain  =     ( -infinity, +infinity)

Range = ( -infinity ....1,2,3]     set builder notation 

Range =  ( -infinity, 3 ]   interval notation

Zeroes   x = -6  ,  x  = -6

Y intercept = occurs at the vertex ⇒  (0,3)

Intervals of increase  = (- infinity, 0)

Intervals of decrease = (0, infinity )

The end behavior in both directions is toward -infinity

Here's the parent graph and the given graph : https://www.desmos.com/calculator/tb3pt4duib

Note that 1.98  can be written as  1 + 98 / 100  = 198/100  = 99/50

Also  note that    

x^2           =           x * x

______             __________

x^2 - 1             (x - 1) ( x + 1)

So we can write

2*2        3*3        4* 4                 (n - 1) ( n -1)       n *  n

____ *  _____  * _____ *  ....... *  ___________  * ___________ <  99 / 50

1 * 3      2 * 4      3 * 5                 (n - 2) ( n)           (n - 1) (n + 1) 

Note that all the terms in red will be "cancelled" in the process and we will be left with

2 n     <    99

____       ___                 multiply both sides by (1/2)   and we have that

(n + 1)      50

  n           <          99

______              ___

(n + 1)                100

And its obvious that the largest integer is   n  =  98