CPhill

4x^3 - 3x^2 + bx + 6  ... we can use the Remainder Theorem, here

So, if he remainders are equal, we have

4(1)^3 - 3(1)^2 + b(1) + 6 = 4(-3)^3 - 3(-3)^2 + b(-3) + 6

4 - 3 + b = -108 - 27 - 3b   simplify

1 + b = -135 - 3b

136 = -4b

b = -34

(2x-3)^6 = 

Let's look at this one again.....

29/289z<16/161z      subtract 16/161z from each side

29/289z - 16/161z  < 0   factor

[ 29/289 - 16/16] z < 0   divide both sides by [ 29/289 - 16/16]

1z < 0    or, just  ....  z < 0

Let H be the height of the larger monument and 5 - H be the height of the smaller.

The exposed surface area of the larger monument is 4H^2 + [H^2 - ( 5-H)^2]  and the exposed surface area of the smaller monument is  5(5-H)^2

And setting this equal to the tottal exposed area, we have

 4H^2 + [H^2 -( 5-H)^2] +  5(5-H)^2 = 61

5H^2 + 4(5 - H)^2  = 61

5H^2 + 4(H^2 - 10H + 25) = 61

5H^2  + 4H^2 - 40H + 39 = 0

9H^2 - 40H + 39 = 0    factor if possible

(9H  - 13)(H - 3) = 0    ...so....the possible solutions are H = 13/9   and H = 3......but, since  H is larger than 5 - H,  the larger monument must have a side of 3 m and the smaller monument must have an edge of 5 - H = 5 - 3  = 2m

LOL!!!.....I'll take the G and T's. too......!!!!

There may be other acceptable answers here, but......I think it's possible that she is 66....

The grandson is 1 when she is 61  ......61 times his age

When he is 2, she is 62.......31 times his age

When he is 3, she is 63..... 21 times his age

When he is 4, she is 64.....16 times his age

When he is 5, she is 65..... 13 times his age

When he is 6, she is 66,.....11 times his age

And that's  six birthdays  !!!    {It's also called....."guess and check'.....LOL!! }

0...believe it, or not  !!!

To see why.......

i^1  = i

i^2  = -1

i^3  = i^2 * i  = -1i = -i

i^4   =  i^2 * i^2  = -1 * -1  = 1

Notice...that when we add "like"  terms....they all "cancel" to 0  !!!! 

Fitst ....count the number of letters in the word =  11

For letters that appear more than once......count the number of times each repeats

A = 3      L = 2    S  = 2    E = 2 

Now.....compute this.....

[ The number of letters in the word folllowed by a factorial sign ]   / [ The number of multiple repeats of each of the letters in the second part above, followed by a factorial sign behind each....we will multiply these together]   ..... so we have....

11! / [ 3! * 2! * 2! * 2! ]   =  831, 600   identifiable "words"

OK.....post some problems you're having trouble with......we'll see what we can do.....!!!