SparklingWater2

do you mean inbetween square root of 12 and square root of 52 or just square root of 12 and 52?

1. 

Pennies:$\dfrac{50}1=\boxed{50}$

Nickels: $\dfrac{2\cdot100}5=\boxed{40}$

Dimes: $\dfrac{5\cdot100}{10}=\boxed{50}$

Quarters: $\dfrac{10\cdot100}{25}=\boxed{40}$

2.

$0.5+2+5+10=\boxed{\$17.5}$

if there are 4 two dollar notes, that means there are 8 dollars the are made up of two dolar notes. so 23-8=15, which is the total of the 5 dollar notes, so 15/5=$\boxed3$ five dollar notes

We're looking for the LCM of 3750, and 6000. the way I would do it is find the GCD of the two numbers, then multiply 3750 and 6000 together and divide it by the GCD:

6000-3750=2250

3750-2250=1500

2250-1500=$\boxed{750}$

$\dfrac{3750\cdot6000}{750}=5\cdot6000=\boxed{30,000 }$ miles

I think you may be over thinking it. Instead of placing the bushes at the very front and the very back, you just put bushes in the very front and leave a gap at the end.

14/$\dfrac78$=16 parts of a mile, so 1/16.

If we want a real number, then ${(x-3)^2-(x-18)^2}$ has to be greater than or equal to $0$, or $(x-18)^2$ has to be less than or equal to $(x-3)^2$

when $(x-3)^2=(x-18)^2$, x has the smallest possible value with real solutions. So we have:

$(x-3)^2=(x-18)^2=x^2+9-6x=x^2+324-36x$

$30x=315$

$x=10.5$

so if x is any less than $\boxed{10.5}$, then the equation $\sqrt{(x - 3)^2 - (x - 18)^2}$ would have no real solutions.

a)  $4x+9y\le200$

b)  $x+y\le15$

c)  $x=4$ and $y=11$

lets call the cost of a pair of socks $s$, and the cost of the T-shirt is $t$. we have:

$2s=t$

$3s+t=100$

we can substitute $t$ for $2s$:

$3s+2s=100=5s$

$s=\dfrac{100}5=20$

$2\cdot20=t=40$

so:

$t+2s=40+40=\boxed{\$80}$

For this, we don't ACTUALLY have to solve for r and s. 

Let us first expand (r-2)(s-2):

$r\cdot{s}+r\cdot{-2}+{-2}\cdot{s}+{-2}\cdot{-2}=rs-2r-2s+4$

if r and s are the roots of y^2 - 19y + 9, then:

$y^2 - 19y + 9 = (y-r)(y-s) = y^2 - (r + s)y + rs$

so r+s=19 and rs=9

going back to the original equation, we have 

$rs-2r-2s+4=9-2(r+s)+4=9-38+4=\boxed{-25}$

if the sum of the two numbers are 36, then we can have

Dice 1 rolls | dice 2 rolls

16               |20

17               |19

18               |18

19               |17

so there are 4 ways.

now the total number of sums possible(if you consider the blank faces numbers)is 20x20=400 possible solutions, so the answer would be 4/400=$\boxed{1/100}$

although if the blank faces are not considered numbers then the total number of possible sums would be 19x19=361, so the answer would be $\boxed{4/361}$