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Number 2) is incomplete in the above answer. 

A square also can end in 4 if its root ends in 8. 

Namely:    8,  18,  28  

So, counting the three that end in 2, there are a total of 6.  

_.

So the correct answer is 16?

would the answer be 11??

2) No number can be squared and have a ones digit of 2 or 3. 

The only way it can have a last digit of 4 is if the number ends in 2 (2, 12, and 22). 

So there are \(\color{brown}\boxed{3}\) numbers that work. 

1) If it is divisible by 7 and 11, it must be divisible by \(\text{lcm}(7, 11) = 77\). 

Can you take it from here?

you forgot to post them...

5) If a 2-digit number is prime, it must end in 1, 3, 7, or 9. 

This means that the only numbers that satisfy the condition will contain 2 of the 4 numbers listed above. 

Try each one out and find the ones that work.

4) Track the last digit: 3, 9, 7, 1, 3, 9, 7, 1, 3 ...

Notice a pattern?

3) If a number has an odd number of divisors, it must be a square number. 

Now, just try each square number until you find the smallest one with 5 factors. 

2) If it is divisible by 12, it must be divisible by 4 and 3. The only cases where it is divisible by 4 are 2, 4, 6, and 8. 

If it is divisible by 3, d must be 1 more than a multiple of 3. The only case that works is 4, so the sum is \(\color{brown}\boxed{4}\)

1) Maximize the x coordinate of the midpoint, and go from there. 

thanks man

Hello Subtonic!

\(The\ solution\ set\ of\ the\ equation\ |x - 1| = 3\ is:\ \ \color{blue}x\in \{-2,4\}\)

  !

you need more information i think

So the answer would be 6?

Well... the people in hell want ice-water, but I’m not going to give them any ...and I’m not going to give you the answer to this question. 

However, the solution is very simple and obvious as snowballs in hell. 

--. .- 

I got 9.

Hey there, Guest!

20-(x + 14) < 9(2x +2) - 15x

Step 1: Simplify both sides of the inequality

−x+6<3x+18

Step 2: Subtract 3x from both sides

−4x+6<18

Step 3: Subtract 6

−4x<12

Step 4: Divide by -4

x>−3

And that's the answer.

Hope this helped! :)

( ﾟдﾟ)つ Bye

4) First, maximize the quarters, then maximize the dimes with what's left, then repeat for the nickels and pennies. 

So, use 3 quarters for 75 cents

Then, use 1 dime to get to 85

Then, use 1 nickel to get to 90

Then, use 3 pennies to get to 93

So \(3 + 3 + 1 + 1 = \color{brown}\boxed{8}\) coins.

3) \(12 = 2^2 \times 3\), \(15 = 3 \times 5\)

This tells us that the number will have 3 prime factors (2, 3, and 5). 

The key here is that \(2^x \times 3^y \times 5^z\) has \((x+1)(y+1)(z+1)\)

The only 3 numbers that multiply to 12 are 2, 2, and 3. 

So, z = 1, y = 1, and x = 2 meaning the number is \(2^2 \times 3^2 \times 5 = \color{brown}\boxed{60}\)

2) There are 6 numbers: 

125

152

215

251

512

521

Just find the prime numbers from here...

1) \({2 \over 9} + {1 \over 7} = 0. \overline{365079}\)

I'll let you work it from here.