All Answers

#1

0

h(x) = f (g(x) = 3 (2x+8) + 4 = 6x +28

inverse is (x-28)/6

GuestNov 21, 2021

#1

+483

+2

If a number is congruent to n modulo n, then it is congruent to (n-n) modulo n, which means it is divisible by n.

Now, (using a calculator because I'm lazy) adding all of these numbers up, we get

1 + 12 + 123 + 1234 + 12345 + 123456 + 1234567 + 12345678 + 123456789 + 1234567890 + 12345678901 + 123456789012 = 137174210008.

Now n must be a factor of 137174210008, and n must be less than 9, so testing, we get that only 2,4, and 8 divide 137174210008 and are less than 9, so $\boxed{n=\{2,4, or \space 8\}}$

Hope this was helpful :)

abcdefghijklmnopqrstNov 21, 2021

#2

+483

+2

Alternative solution:

Pair the numbers up as follows

1	2010
2	2009
3	2008
4	2007
5	2006
...
1004	1007
1005	1006

We can see that there are 2010/2=1005 rows, and the sum of each row is 2011, which is equal to 1 modulo 2010, thus the answer is

1005*1 = 1005

Hope this helped :)

abcdefghijklmnopqrstNov 21, 2021

#1

0

999 mod (9 - 1) ==7 - the 999 count would start from 1 to the left as number 993 (1), 994 (2), 995 (3), 996 (4), 997 (5), 998 (6) and 999 (7).

GuestNov 21, 2021

#1

0

Add 'x' to both sides and add 5 to both sides to get

3x <= 17 divide by 3

x <= 17/3 <====== you can finish I hope .....

GuestNov 21, 2021

#1

+483

+2

We can first factor the denominator as $\frac{1}{(x)(x+2)}$. In order for this to be a terminating decimal, both $x$ and $x+2$ have to only have factors of 2 and/or 5. Testing small values of x, we find that $x = \{2,8, etc.\}$. Looking at this, we can see that the second smallest integer value of x is $\boxed{8}$

.

abcdefghijklmnopqrstNov 21, 2021

#1

+118723

+1

That is correct

MelodyNov 21, 2021

#3

+118723

+1

What is the remainder when 5^(301) is divided by 7?

Let's look for a pattern

5^1 5 mod7 = 5 5^(6n+1) = 5 mod 7

5^2 25 mod 7 = 4 5^(6n+2) = 4 mod 7

5^3 125 mod 7 = 6 5^(6n+3) = 6 mod 7

5^4 625 mod 7 = 2 5^(6n+4) = 2 mod 7

5^5 3125 mod 7 =3 5^(6n+5) = 3 mod 7

5^6 15625 mod 7 = 1 5^(6n) = 1 mod 7

7 78125 mod 7 = 5

8 390625 mod 7 =4

an so we have a pattern

301 = 1 mod 6

so the remainder will be 5

MelodyNov 21, 2021

#1

+678

+1

Umm, this seems like homework in Chemistry, search the question up and you will see that.

Chegg costs money!

Fun Fact: A question from 2015.

Straight

StraightNov 21, 2021

#3

+118723

0

Straight, that definitely was not helpful.

Hartelin was reporting a broken calculator feature.

MelodyNov 21, 2021

#2

+118723

+1

Maybe it has been fixed, it is working alright now.

MelodyNov 21, 2021

#1

+678

+1

Hello Hartelin,

When I use x

Example: x+1=3 = Error: timeout

IT GIVES Error: Timeout

WHY????

HELP PLEASE

That is coming because you type something nonsense or don't know how to do that equation.

Hmm, x+1=3 => x=2, they said by me, but actually, you can calculate this yourself.

But if "Error: timeout" comes, and you want to know the solution for e.g x than you can also use WolframAlpha (I don't want to make advertisement).

Hope that was helpful.^^

Straight

StraightNov 21, 2021

#2

+678

+2

Hello Guest,

there is a calculator, namely: Remainder Calculator | The best Long Division Calculator (calculatored.com)

and the link https://www.calculatored.com/math/algebra/remainder-calculator.

I also got the remainder 5.

Hope that was helpful.^^

Straight

StraightNov 21, 2021

#7

+118723

+1

Of the three-digit integers greater than $500$, how many have two digits that are equal to each other and the remaining digit different from the other two

2 zeros : 4

1 zero : 5*2=10

the rest have no zeros

2 of 1,2,3,4 : 4*5 = 20

2 of 5,6,7,8,9 and one 1,2,3 or 4 : 5*4*2=40

2 of 5,6,7,8,9 and one 5,6,7,8,9 : 5*4*3=60

4+10+20+40+60 = 134

MelodyNov 21, 2021

#1

+1

5^(301) mod 7==5 - the remainder

GuestNov 21, 2021