Rom

$${\frac{{\mathtt{1.4}}}{{\mathtt{5.082}}}} = {\frac{{\mathtt{100}}}{{\mathtt{363}}}} = {\mathtt{0.275\: \!482\: \!093\: \!663\: \!911\: \!8}}$$

$$x^2-16<0$$

$$x^2<16$$

$$-4

assuming everyone has the same chance it's

p=12/900 = 1/75

$$1 \mbox{ billion }=10^9$$

$$(10^9)^{(10^9)} = 10^{9\cdot 10^9}=10^{(9 \mbox{ billion })}$$

that's a 1 with 9 billion zeros after it.

might take a while to write down

assuming interest accrues annually you have

$$FV=PV(1+rate)^T$$

$$FV=9600(1.023)^{14}$$

$${\mathtt{9\,600}}{\mathtt{\,\times\,}}{\left({\mathtt{1.023}}\right)}^{{\mathtt{14}}} = {\mathtt{13\,198.668\: \!008\: \!406\: \!117\: \!257\: \!2}}$$

so you would end up with $13,198.67 after 14 yrs.

the rates at which they can solve problems are combined

$$rate_{Beth}=\dfrac p 2 ~\mbox{problems/hr}$$

$$rate_{Molly}=\dfrac x h~\mbox{problems/hr}$$

$$rate_{both}=\dfrac p 2 + \dfrac x h = \dfrac {p h + 2 x}{2h}$$

the time it takes for both of them to do 400 problems is 400 divided by their combined rate.

$$t_{both}=\dfrac {400}{rate_{both}}=\dfrac {800h}{ph + 2x}$$

$$\dfrac 1 R = \dfrac 1 {r1} + \dfrac 1 {r2} ~mho$$

$$\dfrac 1 R = \dfrac 1 1 + \dfrac 1 3 = \dfrac 4 3~mho$$

$$R = \dfrac 3 4~ohm$$

D=3 yd

R=1.5 yd

$$Area=\pi R^2=\pi (1.5)^2$$

$${\left({\mathtt{1.5}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{7.068\: \!583\: \!470\: \!577\: \!034\: \!8}}$$

so 7.1 sq yds is correct

$$C = \pi D = 3 \pi$$

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{\pi}} = {\mathtt{9.424\: \!777\: \!960\: \!769\: \!379\: \!7}}$$

and 9.4 yds is correct for the circumference

well done

$$I(t)=I_0 e^{-\lambda t}$$

well...

if you go ahead and graph these points they appear to all be on a line.

So take a couple of them and figure out what the slope and intercept is and then double check that they all do lie on that line.

I get a slope of 3 and an intercept of -9 so the line the points lie on is

y=3x - 9

Checking this we get

1 -> -6

8 -> 15

10 -> 21

-20 -> -69

25 -> 66

and you can see that all the points lie on this line.