1 Name: Anonymous 2017-05-17 15:28

Quoted by: >>3,5

1 Name: Anonymous 2017-05-17 15:28

Quoted by: >>3,5

The 5 newest replies are shown below.

Read this thread from the beginning.

Read this thread from the beginning.

3 Name: Anonymous 2017-05-17 15:59

>>1

BODMAS

'brackets','orders','division','multiplication','addition','subtraction'.

this isnt smart, its the most baisic of mathematical algorithm.

6^2 / 2(3) + 4

6^2 / 6 + 4

36 / 6 + 4

*(Post truncated)*

BODMAS

'brackets','orders','division','multiplication','addition','subtraction'.

this isnt smart, its the most baisic of mathematical algorithm.

6^2 / 2(3) + 4

6^2 / 6 + 4

36 / 6 + 4

4 Name: Anonymous 2017-05-17 16:50

((6^2)/(6))+4=10

5 Name: Anonymous 2017-06-11 15:40

6 Name: Anonymous 2018-08-27 00:26

36/2(3)+4

18(3)+4

54+4

58

18(3)+4

54+4

58

7 Name: Anonymous 2018-09-23 18:26

- agno = 0

1 Name: Anonymous 2017-05-26 16:10

I am playing around with some image processing code, and I got stuck on a math problem. Since I am really not that good in math, I can't seem to be able to solve it. The issue is as follows:

a^n + b^n + c^n = d

a, b, c and d are all values larger than 0 but less than 1. I need to find n.

Thanks.

a^n + b^n + c^n = d

a, b, c and d are all values larger than 0 but less than 1. I need to find n.

Thanks.

1 Name: Anonymous 2017-05-14 01:23

I never learned this before and now I need it for my homework.

I need to factor using a special factoring formula but im not sure how to

The question is:

16z^2 -24z+9

Thanks in advance for the help

I need to factor using a special factoring formula but im not sure how to

The question is:

16z^2 -24z+9

Thanks in advance for the help

^{2}รท2(3)+4