All Answers 352713 Answers

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:

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

\({7,7,9,10,22}\)