Imcool

They post as guest mostly because they are just too lazy to login, like all they want is for us to answer their homework and then they go byebye. Their just lazy in general

so treat the 5 scientists as one whole group, so there would be 13 people(The group counts as one). The amount of ways will be 13! right? NO! We need to explore the statement that rotating to match is the same thing, but ill let you figure it out. Hint use the (x y). Its suppose to be vertical so yeah

I first want to point out that the second example was incorrect becuase you need two 2's.

Honestly, I have no idea how to do this problem, Even Cphil is stuck on it: Proof at brainly if you copy paste this question

Well lets do some algebra!

\(a^3 + 3ab^2 = 679\)

\(a^3 - 3ab^2 = 673\)

\(a^3 + 3ab^2 = 679 \) this is equal to

\(a^3 - 3ab^2 = 679 - 6ab^2\)

we can conclude that

\(673 = 679 - 6ab^2\)

\(6 = 6ab^2\)

\(ab^2 = 1\)

Most likely a will be very big because we need a^3 to equal something large since the second term is = to 3.

so a^3 = 676 so a = \(\sqrt[3]{676}\)

b = \(\sqrt{\frac{1}{\sqrt[3]{676}}},-\sqrt{\frac{1}{\sqrt[3]{676}}}\)

a - b = \(\sqrt[3]{676}-\frac{1}{\sqrt{\sqrt[3]{676}}},\sqrt[3]{676}+\frac{1}{\sqrt{\sqrt[3]{676}}}\) that is so definenly incorrect, so plz someone see if this is correct or not

Alright. How many total possiblities are there for a dice rolled twice? 36 choices. (6+5+4+3+2+1)

The amount of possiblities for rolling a 1,2, or 3 

1:

only 1,1 works

2:

1,2 or 2,1 works

3:

1,3 or 3,1 works

total of 5 possible choices out of 36

Your answer is 5/36

The LCM of [1,2,3,4,5,6] is going to be 60. The least 5 digit number that is a multiple of 60 is going to be 10020 = 60 x 170, the greatest being 99960 = 60 x 1666. from 170 to 1666, there will be 1666-170+1 multiples of 60, so 1497 multiples. There is a total of 99999-10000+1 = 90000 5 digit numbers so the odds of a 5 digit number not being divisible by all of the numbers in the set is going to be \(\frac{90000-1497}{90000}\) which is going to be 

NOTE:Guest, the problem says that the number is not divisible by EVERY ELEMENT IN THE SET

\(\frac{29501}{30000}\)

Read your question. it makes absolutley no sense and you have no reason to copy paste either alcumus or ur elementary school homework on this. revise your question