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Very well. thank you.   And how do YOU simplify fractions? 

Do you have some examples with which you need some help?

46,000  (no decimal point)            or    4.6 x 10^4

House 1 is TWICE as tall as house 2.....   House 2 is therefore 1/2 as tall as house 1    1/2 x 24 = 12ft tall

Solo weighs 1/7 as much as Sparky...   or 1/7 of Sparky's weight.

House 2 would be 48 ft because you multiply house 1 by 2 to get house 2.

solo is 7 times lighter than sparky (that all i can say because there is nothing to go off of)

Yea... Hru?

It involves a friend... Thats all...

a little mad. But ill be fine

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Remember  SOHCAHTOA ?    Sin = Opposite/Hypotenuse    You are looking for Hypotenuse in this example and are given angle and Opposite

sin 68 = 18/hyp

Hyp = 18/sin 68   = ~19.4 ft

Got it. Nvm

Let the smallet integer = S
The larger integer = S + 1
1/5[S + S + 1] =S - 64, solve for S
S = 107 - the smaller integer
107 + 1 =108 - the larger integer

http://www.squishable.com/pc/squish_cute_octopus_15/Big_Animals/Squishable+Cute+Octopus

This is the only thing is the whole world that I need want for Christmas. Yes. I'm 17 years old and I want a giant octopus!

Hope everyone has a wonderful holiday season!

Thanks Rom!

Hi Rom, I agreed with you when i first read the question but we interpreted it differently from intended.

This is a simplified school question.  Perhaps we are intended to use calculus, or perhaps not.

I will do it without calculus.

The trajectory of a ball will be parabolic (pre-knowledge on my part).  We are only interested in height and time.

The x does not refer to horizonal distance it refers to time and I am going to change it to a t make it less confusing.

So the ball is traveling in the path of a concave down parabola.   (t,y)     (time, height)

If you let the top of the building be the point (0,0)    that is 0 time and  0 vertical height 

then the ground will be ( 5,-25)

and the maximum height coordinates will be (2,20)

It is a concave down parabola and the t intercepts will be 0 and  4    (becasue it is symmetrical)

so the eqution must be

$y=-a(t-0)(t-4)\\ y=-at(t-4)\\ When \;\;t=2,\;y=20\\ sub\\ 20=-a*2(2-4)\\ 20=-2a*-2\\ 20=4a\\ a=5\\ \text{So if the coordinates are set as I have described, I mean with the top of the buildng being height 0}\\ then\\~\\ y=-5t(t-4) \text{ Will model the path of the ball}$

I hope that helps.

int the prime factorization of 109! what is the exponent of 3?

$\begin{array}{|rcl|} \hline \text{The exponent of $3$ is } \\ && \lfloor \frac{109}{3} \rfloor +\lfloor \frac{109}{9} \rfloor +\lfloor \frac{109}{27} \rfloor +\lfloor \frac{109}{81} \rfloor \\ &=& 36 + 12 +4 +1\\ &=& 53\\ \hline \end{array}$

7. 

As shown in the diagram, points B and D are on different sides of line AC. 

We know that Angle B=2*Angle D=60 degrees and that AC=4sqrt(3). 

What is the distance between the circumcenters of Triangle ABC and Triangle ADC?

line segments:

$\text{Let $AC =4\sqrt{3}$ } \\ \text{Let $AG=GC =2\sqrt{3}$ } \\ \text{Let $AE=EC =r$ } \\ \text{Let $AF=FC =R$ } \\ \text{Let $FE = EG+GF $ }$

The distance between the circumcenters of Triangle ABC and Triangle ADC $= FE$

angle:

$\text{Let $\angle ABC = 60^{\circ} $ } \\ \text{Let $\angle ADC = \frac{\angle ABC}{2} = 30^{\circ} $ } \\ \text{Let $\angle AEC = 2*\angle ABC = 120^{\circ} $ } \\ \text{Let $\angle AFC = 2*\angle ADC = 60^{\circ} $ }$

$\mathbf{EG = \ ?}$

$\begin{array}{|rcll|} \hline AC^2 &=& 2r^2\Big(1-\cos(120^{\circ})\Big) \quad \text{$\cos$-rule} \quad | \quad \cos(120)^{\circ} = -0.5 \\ (4\sqrt{3})^2 &=& 2r^2(1+0.5) \\ 48 &=& 2r^2(1.5) \\ 24 &=& r^2(1.5) \\ r^2 &=& 16 \\\\ AG^2 + EG^2 &=& r^2 \\ (2\sqrt{3})^2 + EG^2 &=& 16 \\ 12 + EG^2 &=& 16 \\ EG^2 &=& 4 \\ \mathbf{EG} & \mathbf{=}& \mathbf{2} \\ \hline \end{array}$

$\mathbf{GF = \ ?}$

$\begin{array}{|rcll|} \hline AC^2 &=& 2R^2\Big(1-\cos(60^{\circ})\Big) \quad \text{$\cos$-rule} \quad | \quad \cos(60)^{\circ} = 0.5 \\ (4\sqrt{3})^2 &=& 2R^2(1-0.5) \\ 48 &=& 2R^2(0.5) \\ 24 &=& R^2(0.5) \\ R^2 &=& 48 \\\\ AG^2 + GF^2 &=& R^2 \\ (2\sqrt{3})^2 + GF^2 &=& 48 \\ 12 + GF^2 &=& 48 \\ GF^2 &=& 36 \\ \mathbf{GF} & \mathbf{=}& \mathbf{6} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline FE &=& EG+GF \\ &=& 2+6 \\ \mathbf{FE} & \mathbf{=} & \mathbf{8} \\ \hline \end{array}$

The distance between the circumcenters of Triangle ABC and Triangle ADC is 8

George is about to get a certain amount of change less than one dollar from the cash register.

If he gets the most quarters possible and the rest in pennies, he would need to receive 3 pennies to meet the amount.

If he gets the most dimes possible and the rest in pennies, he would need to receive 8 pennies to meet the amount.

What is the sum, in cents, of the possible amounts of change that he is trying to get?

$\text{Let $n =$ quarters}\\ \text{Let $m =$ dimes}$

Formula:

$\begin{array}{|rclrcl|} \hline \mathbf{25n+3} &=& \mathbf{10m+8} \\ 10m &=& 25n-5 \\\\ m &=& \dfrac{25n-5}{10} \\\\ &=& \dfrac{20n+5n-5}{10} \\\\ &=& \dfrac{20n}{10}+\dfrac{5n-5}{10} ~|~ \text{cut off whole parts} \\\\ m &=& 2n+ \underbrace{\dfrac{5n-5}{10}}_{=a} \qquad m = 2n+a \qquad (1) \\\\ & a &= \dfrac{5n-5}{10} \\\\ &10a &= 5n-5 \\\\ &5n &= 10a+5 \quad | \quad :5 \\\\ &\mathbf{n} & \mathbf{=} \mathbf{2a+1} \qquad (2) \\\\ m &=& 2n +a \\ &=& 2(2a+1) +a \\\\ \mathbf{m} & \mathbf{=}& \mathbf{5a+2} \qquad (3) \\ \hline \end{array}$

conclusion:

$\begin{array}{|rcll|} \hline \mathbf{n} & \mathbf{=}& \mathbf{2a+1} \qquad a \in \mathbb{Z} \\ \mathbf{m} & \mathbf{=}& \mathbf{5a+2} \qquad a \in \mathbb{Z} \\ \hline \end{array}$

solution:

$\begin{array}{|c|r|r|r|} \hline a & \text{quarters} & \text{dimes} & \text{pennies} \\ \hline 0 & 1 & 2& 28 \\ 1 & 3 & 7& 78 \\ 2 & 5 & 12 & 128 > $1~ ! \\ \hline \end{array}$

${7,7,9,10,22}$