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The calculator is on the home page.  

(5-2)^2-(6-5)^3 =

3^2  - 1^3  =

9 - 1  =

8

(x^4+2x^3-7)÷(x^2-1) =

                x^2  + 2x  + 1

             --------------------------------------

x^2 -1   [  x^4  + 2x^3  + 0x^2   + 0x  - 7 ]

                x^4               -  x^2

------------------------------------------------

                           2x^3  +   x^2

                            2x^3               -2x

---------------------------------------------------

                                          x^2  + 2x - 7

                                          x^2          - 1

                                        ----------------

                                                    2x - 6

So, our answer is  x^2 + 2x + 1   +  [2x -6] / [x^2 -1]

First, let us show that it is true for n=1

So....a^(2(1) + 1) + b^(2(1) + 1)  =  a^3 + b^3  =  (a +b)(a^2 -ab + b^2)...and (a + b) is a factor

Now, let us assume it is true for  k=1, that is, a^(2(k) + 1) + b^(2(k) + 1) is true

Now, let's prove that it is true for k + 1

So we have

a^(2(k+1)+ 1) + b^(2(k+1) + 1) =

a^(2k + 3) + b^(2k + 3) =

a^3*a^(2k)  + b^3*b^(2k)......and using   a^3  =  (a +b)(a^2 -ab + b^2) - b^3 , we can write

[(a+b)(a^2 -ab + b^2) - b^3]a^(2k)  + b^3*b^(2k)  =

(a+b)(a^2 -ab + b^2)a^(2k) - b^3[ a^(2k) -  b^(2k)]

Notice that the last term is a difference of two even powers.........and it can be shown that the difference of two even powers can be factored with (a + b) as a factor thusly...

(a^(2n) - b^(2n)) = (a + b)[a^(2n-1) - a^(2n-2)b + a^(2n-3)b^2 - a^(2n-4)b^3 +.....+ ab^(2n-2) - b^(2n-1) ]

Thus, (a+b) is a factor of both terms, so (a +b) is a factor of a^(2k + 3) + b^(2k + 3)

12i³m + 12i⁴

If      $${i}$$ =   $${\sqrt{-{\mathtt{1}}}}$$

then this simplifieds to 

12*-1im+12*-1*-1 = -12mi+12

You have to move that point one more time

9.5% = 0.095

$${\mathtt{0.095}}{\mathtt{\,\times\,}}{\mathtt{40}} = {\frac{{\mathtt{19}}}{{\mathtt{5}}}} = {\mathtt{3.8}}$$

3.8cm

Yes we do miss you Kitty.  :(

Did you read my New Year Day's Post?

Could  the person who posted this please insert brackets to clarify the meaning?

Chris has asked me to look at this one and YES Chris I got the same answer.

I did it almost the same the only minor difference was that I repositioned the origin from the ground to the bottom of the needle.  I doubt that it made it any easier though.

It is all too much for me!

I don't want to mess with that chimp!         LOL  

Using Calculus, we can find the derivatve of the function.... which is the velocity at any time t..so we have

h' =-32t + v₀

And at the max ht, this is = 0, so we have

-32t + v₀ = 0     →  v₀ = 32t  ...and, substituting this back into the position function at the time of the max ht, we can find the time it takes to reach that point...so we have

1081 = -16t^2 + (32t)t + 921    simplify

1081 = -16t^2 + 32t^2 + 921

160 = 16t^2    divide both sides by 32

10 = t^2

t = √10 sec  ..and this is the time it takes to reach the max ht

So, using the first derivative to solve for v₀ when t= √10  , we have ... v₀  = 32(√10)ft/s

And our original position function becomes

h = -16t^2 + 32√10t + 921

217,772,275.41

100km/hour = 100000m/3600sec

You have t=10, u=0, v=100 and you need to find s

the only equation that has all of those letters and no others is [3]

100km

$$s=\frac{1}{2}(0+\frac{100000}{3600})*10 \;metres$$

$${\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{100\,000}}}{{\mathtt{3\,600}}}}\right){\mathtt{\,\times\,}}{\mathtt{10}} = {\frac{{\mathtt{1\,250}}}{{\mathtt{9}}}} = {\mathtt{138.888\: \!888\: \!888\: \!888\: \!888\: \!9}}$$

The car will go 139 metres (to the nearest metre)

Let the base be representd by h-8

And the area is given by  b*h /2 = (h-8)*h / 2

And this is "at least" 60cm² ....and "at least" is analogous to "≥" ... so we have...

(h-8)*h / 2  ≥  60 cm²   .... which we could also express as....

h² - 8h  ≥ 120 cm²

Remember that  (1 / aⁿ )  = a^-n

Since 216 = 6³ , we have (1/216)  =(1 / 6³ )  = 6^-3

And 100,000 = 10⁵ , so, 1/100,000 = (1 / 10⁵ )  = 10 ^-5

And one millionth  = 1 / 1,000,000 = 1 / 10⁶ = 10^-6

The "formula" for the sum of an infinite sequence is given by

S = a₀ / (1 - r)....where S is the sum, a₀ is the initial value, and r is the common ratio...so we have...

6 = 2 / (1 - r)     mutiply both sides by 1-r

6(1 - r)  = 2        simplify

6 - 6r  = 2           rearrange

4 = 6r                solve for r

r = 2/3

Haha thanks, CPhill! Happy New Year to you guys too! :D

12 in x 12 in  = 144 sq in

And, assuming 19,836 is in sq in, we have

19,836 / 144  = about 137.75 pieces....(and assuming that we have to have "whole" pieces, let's round this up to 138.....)

Well, 6 months ago, he must have owed 6 x 12.65 more than he does now = $75.90

So $75.90 + $132.50 = $208.40 ......just as "Anonymous" found.....!!!

Converting 6.71 x 10⁸ ,   we just move the decimal point 8 places to the left...so we have

671,000,000 mph

And if you wanted to find out how long it takes light from the Sun to reach Mars, we have

[ 1.42 x 10⁸]miles / [ 6.71 x 10⁸ ] mph =  about .2 hrs, or about 12 minutes

1 1/3 / 1/4  x  2/5 / 1/2

Remember that 1+1/3  = 4/3...so we have

4/3 / 1/4  x   2/5  / 1/2     and when we divide fractions like this, we just keep the munerator as it is, change to multiplication, and  "flip" the fraction in the denominator...so we have

[4/3 X 4/1 ]  x [ 2/5 x 2/1] = 

16/3  x  4/5 =

64/15

And that's ir...

208.4

it the same as doing 73-19

The value of the cosine must be a fraction no smaller than -1.000 nor no larger than +1.000. So a cosine value of 116.07 is impossible.

Percent means "per hundred"

5% = 5/100 

5% of something means 5% * something.

So, we get:

5% * 125 000 = 5/100 * 125 000 =  6250