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Since there are 6 side on each die, and you are rolling two, there are 36 possibilities (6 faces*6 faces).

You can roll a sum of 5 with:

1, 4

2, 3

3, 2

4, 1

There are 4 ways.

That means 4 out of the 36 ways you can roll two dice (1/9 once rounded). That means there is a 1/9 chance of rolling a total of 5.

As for rolling a number greater than 3, you can roll 4, 5, or 6 on each die, which means there are 6 ways to roll two dice and get a number greater than 3 on at least one. This means that there is a 6/36 (1/6 once rounded) chance to do the same.

In conclusion, there is a 1/9 chance of getting a total of 5 and a 1/6 chance of rolling at least one number greater than 3 when rolling two dice.

Very nice, Anonymous4338..........5 pts from me.....!!!

Very kind thank you geno3141 and heureka for taking your time to help us!

We are digesting!  Have a nice day!

We can divide this up a bit.

2x+13<6x+1

AND

6x+1<4x+11

First, solve 2x+13<6x+1.

2x+13<6x+1

Subtract 1 from each side...

2x+12<6x

Subtract 2x from each side...

12<6x

Divide by 6 on each side...

2<x

There you go! Now, let's do 6x+1<4x+11.

6x+1<4x+11

Subtract 1 from each side...

6x<4x+10

Subtract 4x from each side...

2x<10

Divide by 2 on each side...

x<5

Okay! Now we know that 2<x and x<5.

In conclusion, 2<x<5.

Could you clarify a bit? The post is definitely working; everyone can see it, but nobody understands what you're trying to say.

Add

verb (used with object)

1.to unite or join so as to increase the number,quantity, size, or importance:

to add two cups of sugar; to add a postscript toher letter; to add insult to injury.

2.to find the sum of (often followed by up): 

Add this column of figures. Add up the grocerybills.

Very nice, heureka and Bertie......!!!!!

what is the amount for rs.5000 at 12% compounded bianually

FV=PV(1+r)^n

FV=5,000(1 + 12/200)^2

FV=5,000(1.06)^2

FV+5,000 X 1.1236

FV=5,618.00- for one year.

I realize that Melody, but like so many other posts on here, you have to make some assumptions about what is being asked...Often I will write my assumptions in my posted answers, but this time I made the assumption that =0 was just left off in the question and solved it with that....

As Melody pointed out the other day, the Web2.0 calculator can tell you if a number is prime

Enter   'factor'  then your number in question (in parentheses), then '='       If it only comes up with your number as the  answer, it is prime....try it!

O_o

Yep.....   What's up?

Vey nice, heureka......!!!!!

What's the question????

6x + 23= 67

6×=67-23

6×=44

×=44/6

×=7.3333....

A prime number is a whole number (counting number, natural number) that is larger than 1 but only divisible by itself and 1.

Some prime numbers are 2, 3, 5, 7, 11, ..

4 is not a prime number because it is the product of 2 x 2 as well as 1 x 4.

6 is not a prime number because it is the produt of 2 x 3 as well as 1 x 6.

2 is a prime number because the only product is 1 x 2.

3 is a prime number because the only product is 1 x 3.

If the number is a prime number it can only be factored into 1 and itself.

A number is not a prime number if it can also be factored in a different way besides 1 and itself.

Hello,

I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!

Let f(x)= ax / (2x+3)

Investigate if you can define the 'a' so that f(f(x))=x

\(\begin{array}{rcl} f(x) &=& \frac{ax}{2x+3} \\\\ f(f(x)) &=& f(\frac{ax}{2x+3}) =x\\\\ f(\frac{ax}{2x+3}) &=& \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} = x \\\\ \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} &=& x \\\\ a\left( \frac{ax}{2x+3} \right) &=& x \left[ 2\left( \frac{ax}{2x+3} \right) +3 \right] \\\\ \frac{a^2}{2x+3} &=& 2\left( \frac{ax}{2x+3} \right) +3 \qquad | \qquad \cdot (2x+3)\\\\ a^2 &=& 2ax +3(2x+3) \\ a^2 &=& 2ax +6x+9 \\ a^2-2x\cdot a -6x-9 &=& 0 \\\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ a_{1,2} &=& {2x \pm \sqrt{(2x)^2-4(-6x-9)} \over 2} \\ a_{1,2} &=& {2x \pm \sqrt{4x^2+4(6x+9)} \over 2}\\ a_{1,2} &=& {2x \pm 2\sqrt{x^2+6x+9} \over 2} \\ a_{1,2} &=& x \pm \sqrt{x^2+6x+9} \\ a_{1,2} &=& x \pm \sqrt{(x+3)^2} \\ a_{1,2} &=& x \pm (x+3) \\ a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ 3 } \end{array}\)

Hi Melody: Thanks for going to all that trouble calculating this problem in detail. However, there is a much easier method of doing it, which you can make note of for future references. It goes something like this:

Rate of investment 4%

Rate of increase in payment, or inflation rate of 3%, therefore the net investment rate is:

1.04/1.03=0.9708737...%. Then we simply use the standard TVM formula of finding the PV of 20 payments of $2,000 each, which is:PV=P[((1 + r)^n-1) / ((1 + r)^n.r)], which gives the following result: PV=2,000[((1.009708737)^20-1) / (1.009708737)^20* .009708737)], which gives:

PV=2,000 X 18.0985842...=$36,197.17. But this amount already includes 1.03 increase in the first  payment. Therefore the PV=$36,197.17 / 1.03=$35,142.88.

Since  f(x) = ax/(2x+3),  to find  f( f(x) ), replace the x=term in  ax/(2x+3)  wwith  ax/2x+3):

f( f(x) )  =  f( ax/(2x+3) )  =  [ a( ax/(2x+3) ) ] / [ 2( ax/(2x+3) ) + 3 ]

    =   [ a²x / (2x+3) ] / [ 2ax / (2x+3) + 3]

Assume that  x  isn't  -3/2:  multiply both the numerator and denominator by  2x + 3 and simplify:

    =  a²x / ( 2ax + 3(2x+3) ]

    =  a²x / ( 2ax + 6x + 9 ]

Since f( f(x) ) = x:     a²x / ( 2ax + 6x + 9 ]  =  x

Multiply both sides by  2ax + 6x + 9:     a²x  =  x(2ax + 6x + 9)

Assuming that x isn't 0, divide both sides by x:     a²  =  2ax + 6x + 9

--->     2ax + 6x  =  a² - 9

--->    2x(a + 3)  =  (a + 3)(a - 3)

If a is not equal to -3, divide both sides by  a + 3:

--->   2x  =  a - 3

and    x  =  (a - 3) / 2                        

But, x could be zero (it checks in the original expression).

Also checking  -3/2  and  -3 -- they don't check.

So, the answer:  either  0  or  (a - 3)/2.

If n is a positive integer such that 2n + 1 is a perfect square, show that n + 1 is the sum of two successive perfect squares.

2n + 1 is a perfect square:

\(\begin{array}{rcl} 2n+1 &=& a^2 \\ 2n &=& a^2 - 1 \\ n &=& \frac{a^2-1}{2}\\ \hline n+1 &=& \frac{a^2-1}{2} +1 \\ n+1 &=& \frac{a^2-1+2}{2}\\ n+1 &=& \frac{a^2+1}{2}\\ \end{array}\)

Because \(\frac{a^2+1}{2}\) is a integer then \(a^2+1\) is even and divisible by 2, then \(a^2\) is odd or also \(a\) is odd.

A odd number is  \(2b+1\)

\(\begin{array}{rcl} n+1 &=& \frac{a^2+1}{2}\qquad \text{substitute }\ a = 2b+1\\ n+1 &=& \frac{(2b+1)^2+1}{2} \\ n+1 &=& \frac{4b^2+4b+1+1}{2} \\ n+1 &=& \frac{4b^2+4b+2}{2} \\ n+1 &=& 2b^2+2b+1 \\ n+1 &=& b^2 + b^2+2b+1 \\ n+1 &=& b^2 +(b+1)^2\\ \end{array}\)

So n + 1 is the sum of two successive perfect squares and \(b = \frac{a-1}{2}\).

Example:

\(\begin{array}{rcl} a &=& 5\\ 2n+1 =a^2 &=& 25 \qquad \rightarrow\qquad n = \frac{a^2-1}{2} = \frac{25-1}{2} = 12\\\\ b &= &\frac{a-1}{2} = \frac{5-1}{2} = 2\\ n+1 &=& 12+1 = 13 \\ 13 &=& b^2 + (b+1)^2 = 2^2+3^2 = 4+9\\ \end{array}\)

XD...

Hi, adylyn_rae!

Could you clarify, please?

If you just mean to encourage our users to help people on the forum, then thanks! We'll make sure to do so.

If you want to thank somebody, I think it would helpful if you included their name, or if you just posted on your own question.

It's really thoughtful of you to thank whoever answered your question!

6x+23=67

Let's go through this step by step.

First, subtract (take away) 23 from both sides...

6x=44

Then, divide by 6 on each side.

x=7 1/3

OR

x=7.33