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Please put your second question in a different post. (And delete it from here)

I have not seen that symbol before in latex.  It is interesting seeing new things. Thanks guys.

D. 9 + (-3)

B. distance

Although "phone numbers" is a rather interesting  choice....LOL!!!!

As long as we're not  specifying an ordered set, order does not  matter    {1,2,3}   is the same set as { 2,3,1 }

False.....whole numbers are just the  positive integers (the "counting  numbers" ) 1,2,3,4,5.....etc.

B. False

False.....2 is the only even prime...all the  rest are  odd

See here : https://web2.0calc.com/questions/is-this-statement-true-or-false_11

The whole  numbers are  generally  considered  to be the  "counting" numbers  1,2,3,4,5.....

So....they don't  include 0....so.....B

I'm assuming this is : 

4^(2x) * 1/16 = 4^(6x+18)

Note that 1/16  =   1 / 4^2 =    4^(-2)   ....so we have

4^(2x) * 4^(-2)  = 4^(6x + 18)           using a property of  exponents, we can write

4^(2x  -2)  = 4^(6x + 18)        we have the  bases the  same.....we can solve  for the  exponents

2x  - 2 =  6x  + 18       add 2 to both sides....subtract 6x from both  sides

-4x = 20      divide both sides  by  -4

x  = -5

this has been answered by CPhill on a different quetsion here is the link

https://web2.0calc.com/questions/need-this-quick_1

Hope this helps!

This statement is false; 2 is a prime and even number.

Thanks! :)

\(\text{let $t=4+f,~f\in [0,1)$}\\~\\ 4 \leq (4+f)+f < 5\\ 0 \leq 2f < 1\\ 0\leq f < \dfrac 1 2\\~\\ 5 \leq 4+f + 2f < 6\\ 1 \leq 3f < 2\\ \dfrac 1 3 \leq f < \dfrac 2 3\\~\\ \text{both of these inequalities must be met}\\ \dfrac 1 3 \leq f < \dfrac 1 2\\ \dfrac{13}{3} \leq t < \dfrac 9 2\)

Thankyou

\(\text{It's probably easiest to derive the log of this product and the take the exponential of the log.}\\ p = \prod \limits_{k=1}^\infty \left(2^k\right)^{\frac{1}{3^k}}= \prod \limits_{k=1}^\infty 2^{\frac{k}{3^k}}\\ \log(p) = \sum \limits_{k=1}^\infty \log\left(2^{\frac{k}{3^k}}\right) =\log(2) \sum \limits_{k=1}^\infty\dfrac{k}{3^k} = \\ \log(2) \dfrac{3}{(3-1)^2} = \log(2) \dfrac 3 4\\ e^{\log(p)} = e^{\log(2)\frac 3 4} = 2^{\frac 3 4}\)

never mind I thought that the question was asking something different I got the correct answer. thanks for pointing that out for me. 

I know but the question is asking all of the numbers between the inequality I don't know if i counted them all correctly

You really ought to be able to glance at that inequality (the bottom one) and immediately see the two integers that satisfy it.

It's just an expression, don't take it literally.

You need to look at Q1 as well.

any day at an office would be a bad day.

f(x)  =  √[x^2]

Note, Logic,  that   √[x^2]    = l x  l

And the range of   l x l    = [ 0, inf)

Well....welcome aboard!!!

Note....you could register and  become a member.....at no charge  !!!!

Also I have never even used a website like this before.