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actually (infinity)+(infinity)=(infinity) because "infinity" doesn't have a number or "exact amount" 

done

but  you were a good guy before and i seen that under all of your act you weren't a bad guy

Just because this board has the off-topic flag doesn't mean that it's facebook or reddit.

This site is here to allow folks to get "help" with their math problems. 

There are a zillion other sites out there for idle chatter.

Use one of them.

What about me to be honest I think I've changed quite a bit...I mean am I wrong

Here's an alternative approach:

You could try using WolframAlpha to factor your large integers. For example:

oh wow i didnt know you can do binomial expansion with algebraic number

thank you

oh wow i didnt know you can do binomial expansion with algebraic power

thank you

perfect! i forgot about that haha thank you

b)    Show that this is true for n = 2

15^2 - 8^(2 - 2)   =  225  - 1  =  224   = 32(7)

Assume it is true for n =  k.....that is

15^k - 8^(k - 2)    is a multiple of 7

Prove it is true for  k + 1

15^(k + 1)  - 8^(k + 1 - 2)

15^(k + 1) - 8^( k - 1)    note YEEEEEET  that we can write this as

(14 + 1)^(k + 1) - (7 + 1)^(k - 1)          use the binomial theorem and expand

[14^(k + 1) + C(k + 1, 1)*14^k  + ... + 14 + 14^0* 1^(k + 1) ] -

[ 7^(k - 1 )  + C(k - 1, 1) * 7^(k - 2) + ....+7 + 7^0*1^ (k - 1) ]   

The  terms   14^0*1^(k +1)   and  7^0*1^(k - 1)    just equal 1  and will "cancel" with the subtraction of the second expansion from the first

So...we have this simplification.... 

[ 14^(k + 1) + C(k + 1, 1)* 14^k  +...+ 14  - 7^(k - 1) - C(k - 1, 1)*7^(k - 2 ) -...- 7 ]  =

[ (7 *2)^(k + 1) + C( k + 1,1)* (7*2)^k + ....+ 14 - 7^( k - 1) - C(k - 1, 1)*7^(k - 2) -....- 7 ]

Every term in the expansion simplification will be divisible by 7, hence the result is a multiple of 7

"can you explain why you did Ur + U n+1 at the first part of ii please "

Think of the summation as follows:  \(\sum_{k=1}^nu_k=u_1+u_2+...+u_n\)

Hence \(\sum_{k=1}^{n+1}u_k=u_1+u_2+...+u_n+u_{n+1}\)  which is \(\sum_{k=1}^{n+1}u_k=\sum_{k=1}^nu_k+u_{n+1}\)

Hope that makes it clear.

thank you for the detailed explanation

can you explain why you did Ur + U n+1 at the first part of ii please its the only part i didnt catch on to

Solution for part b:

.

ah thank you very much

YA YEEET

Let the volume be

V =  x^2*h      where x is the side of the base and h is the height    (1)

And.....using the total cost for the surface area....we know that

Total cost for top/ bottom + Total cost for sides =  15

2x^2(3) + 4xh(1.5) =  15

15 - 6x^2  = 6xh

[ 15 - 6x^2] / [6x]  = h      (2)

Put (2)  into (1)  and we have that

V = x^2 [ 15 - 6x^2] / [ 6x]

V =  x [ 15 - 6x^2] / 6

V =  (1/6) (15x - 6x^3 )       take the derivative and set to 0

(1/6) (15 - 18x^2) = 0

18x^2 =  15

x^2  =  15/18    =  5/6

x = √[5/6]  

So.....the volume of the box is :    (1/6)√[5/6] (15 -6 (5/6) )  = (1/6)√[5/6] (10) =  (5/3)√[5/6] ft^3    

Thanks Chris,

I would not have thought to do that. :))

LOL!!!.....not possible.....!!!!

1. When 310 is divided by a two-digit positive integer K, the remainder is 37. What is the sum of all possible values of K?

310 - 37 =  273

273 = 3  * 7 * 13

310 / 39 =  7(39) + 37

310 / 91 = 3(91) + 37

Sum of all possible K's  =   39 + 91  =  130

so it's best if i think in terms of \(x\) when facing a quadratic trig/similar trig equations?