CPhill

how do i write 14400000 inscientific notation

1.44 x 10^7

5^-4 =

1 / (5^4) =

1 / (5*5*5*5) =

1 / 625

There you go !!

answer to 3y2-2y-8

I assume you want to facor this...it's just

(3y + 4) (y - 2 )

That's it !!!

Depends on what kind of pie we have.......

what is 4^6*x^8y^-2/ 4^7*x^4y^2

So we have

[(4^6)*(x^8)*(y^-2)] / [(4^7) * (x^4) 8 (y^2)]

So, by the property of exponents that says that  (a^n) /  (a^m)   =  a^(n-m)  We can write

[4^(6-7)] * [x^ (8-4)] * [y^(-2-2)]  =

[4^(-1)] * [x^4] * [y^(-4)] =

[x^4] / [4 * y^(4)]

simplify (64)x^9)^(1/3)

OK...here's what we really have

64^(1/3) * (x^9)^(1/3) =

4 * (x) ^( 9 * (1/3)) =

4 * x^(3) =

4x^3

And that's all......

Re: Factorise fully: (x^2-16) - (x-4)(3x+5)

OK...this one isn't too difficult.....note that the first thing is just a difference of two squares, so we can write it as (x+4) (x-4)   

And we have

(x+4) (x-4)  - (x-4) (3x-5)

Since (x-4) is common to both terms, we can factor it out, and this gives us

(x-4) [(x+4) - (3x -5)]  Now, simplifying the stuff in the brackets, we have

(x-4) (9 - 2x)

And that's it !!!

-2log8(x + 1) = -8

It's a little difficult to tell what you're using as the base ....if it's 10, we have this

Let's divide both sides by -2   This gives

log8(x + 1) = 4

Writing in exponential form, we have

10^4 = 8 (x + 1)   Therefore

10,000 = 8(x + 1)   Now, divide both sides by 8  This gives

1250 = x +1      Subtracting 1 from both sides gives us

x = 1249

However, if you're using 8 as the base, let's take it from this point

log8(x + 1) = 4

Now, let's write this in exponential form using base 8......we have

8^4 = x +1

4096 = x + 1    And subtracting 1 from each side gives us

x = 4095

I suspect that the second thing might be what you meant, but anyway, you have answers using both bases!!!   

Re: What is the sum of the first 55 consecutive odd numbers?

Mmmmmm....this is an interesting one !!  Let's see if we can detect a "pattern"

1 = 1

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 +7 =  16

Anything "jump out" at you ??

Note that the first one is just 1^2

The second is just 2^2

Thr third one is just 3^2

So...it appears that the sum of the first n consecutive odd numbers is just n^2

I'll let you take it from there......

BTW...for "extra credit"....see if you could find a pattern for the sum of the first n consecutive even numbers!!

(1/1+r)30/r-100=0

So we have

30 / [(1 +r ) (r )] - 100 = 0

Add 100 to both sides...this gives

30 / [(1 +r ) (r )]  = 100   Now multiply both sides by [(1 +r ) (r )]   This gives us

30 = 100 [(1 +r ) (r )] Let's do some simplification on the right side

30 = 100r (1 + r)

30 = 100r + 100r^2 

Now, let's divide everything by 10 to reduce the numbers

3 = 10r + 10r^2     And rearranging things a bit, we have

10r^2 + 10r - 3 = 0

This won't factor. so using the on-site solver, we have

{r=-(((sqrt(55)+5)/10)), r=((sqrt(55)-5)/10)}

So r has two values

The first one ≈ -1.24

And the second one ≈ .242

Not very nice and neat, but that's it