All Answers 358160 Answers

These are two different things....

Here are the "formulas" for the frustum volumes of various objects.....

The geometric mean of N numbers is just the Nth root of their product.....so, since we have two numbers, we have....

 √(10 * 30) = √300 = 10√3

Melody , thank you for your help!!!!!!!!

1. Energy is a scalar, not a vector quantity, so you would add not subtract the energies.

2. If it were an elastic collision they wouldn't stick together.

3. It would be unusual not to lose a lot of energy (dissipated through frictional heat loss, for example) in a head-on car crash! 

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Let N = number enrolled

N- N/12 present or 11N/12 present

11N/(12*5) on field trip, so 4*11*N/(12*5) in school or 11N/15 in school

11N/15 = 704

N = 15*704/11

N = 960

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A question for Alan or Heureka

Why does this equation set fail? Would this work if this were an elastic collision?

Kinetic energy autoN = (0.5(1300)*(30²)) = 585,000J

Kinetic energy autoS = (0.5(900)*(15²)) = 101,250J

585,000J + (- 101,250J) = 483,750J

Total mass now as a unit = (1300kg) + (900kg) = 2200kg

Solve for velocity: Sqr(483750J / 0.5(2200kg)) = 20.97m/s in north direction.

_7UP_

Can't you ever be polite Nauseated.  

Don't let him worry you Sorasyn, Nauseated calls himself Nauseated because he is always Nauseating!

No problem SevenUP!

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Sorry Alan, I didn't see your post until I made mine.

HHAHA!!!...I wasn't reading the question that was presented and I started answering part A..I'll deduct that 3....(maybe!!!)

The answer to your question is yes.

A positively charged rod will pull electrons away from the ball until it reaches an equilibrium. This will leave the ball with a net positive charge.

_7UP_

Yes,  if the rod was positively charged the ball in step a will be charged positively.

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Hey Chris, how come you got 3 points for an empty answer??   LOL

I want some of those too.    

I wanted to do this answer exact but it is getting all too difficult and I don't know if what I have done is correct anyway. 

$$\\ED=1.5h=1.5*112\approx 168\\
KD=\sqrt3*h = \sqrt3*112\approx 194\\
\triangle AKD \equiv \triangle HKD \;\;\; $Three equal side test$\\\\
so\;\; KDA=30^0\\\\
But $Now consider triangle EKD$\\
EK^2=168^2+194^2-2*168*194*cos(78.2)$$

$${\sqrt{{{\mathtt{168}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{194}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{168}}{\mathtt{\,\times\,}}{\mathtt{194}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{78.2}}^\circ\right)}}} = {\mathtt{229.194\: \!522\: \!964\: \!077\: \!551\: \!3}}$$

$$\\EK \approx 229.2\\\\
Let \;\;\alpha $ be the angle of elevation sort, that is, $ tan\alpha = \frac{HK}{EK}\\\\
tan\alpha = \frac{h}{229.2}\\\\
tan\alpha = \frac{112}{229.2}\\\\
\alpha=tan^{-1}\frac{112}{229.2}\\\\$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{112}}}{{\mathtt{229.2}}}}\right)} = {\mathtt{26.042\: \!733\: \!855\: \!708^{\circ}}}$$

So if i have not made any stupid mistakes (which I probably did)  then  the minimum angle of elevation is approx 26 degrees and it is from the point E

Melody is correct.....the minimum angle of elevation, from John's perespective, will occur at E, since this is the greatest distance from "h."

To find this angle of elevation, we need to know what EK is

Melody has already calculated angle E = cos^-1 (1/9) = about 83.62°

And since triangle AED is isosceles, angle DAE = (180 - 83.62)/2 = 48.19°

And angle KAD can be found using the Law of Sines, thusly

sin KAD/KD = sin90/AD    and KD = 112/tan30  and AD =2h = 224

So we have

sin KAD =(112/tan30) /224   so sin^-1 ((112/tan30)/224) = 60°

So angle DAE + KAD = angle KAE = 48.19 + 60 = 108.19°

And using the Law of Cosines, we have

EK^2 = AK^2 + AE^2 -2(AK)(AE)cos(108.19)

EK^2 = (112/tan30)^2 + 168^2 - 2(112/tan30)(168)cos(108.19)

EK = about 293.6

And the angle of elevation from E is given by

tan^-1(112/293.6) = about 20.88°

Edit ....I have given the wrong measure for KA...it should be 112/tan 45 = 112

So we have

EK^2 = (112)^2 + 168^2 - 2(112)(168)cos(108.19)

EK = 229.162

And now my angle of elevation will = Melody's answer...!!!!  {more or less}

Well, you’ve been a way for two months. Did you seek treatment for your CDD? Remember, it takes two months just for the therapists to convince you that you have it.

Anyway, do you have more of the special mixture of NaClO₃ that produces ozone? I’m constantly in hostile environments and I’ve used most of the last batch.

This is a pity. I was hoping you would come for dinner after you were bigger. I made a special camelid sauce called dromedary hump sauce, just for you. The inspiration came from the piccolo sauce that went well with oboeduck. (Oboduck is out of season, now).I do hope you get better; if not, the invitation extends to your brother or sister, of course.

I hope your doctor will walk a mile for a camel. Your symptoms indicate one of the equine viruses, or Coccidioidomycosis, or Clostridium Welchii. If it is C. Welchii, and you recover, we’ll have to wait about 3 extra months for the invitation. There is a small boat load of other pathogens included too, but none of them include llama flu; there is no such thing.

In any case, I really do hope you get better.

...................................

7*3 = 21

Seven by itself is seven (7*1=7), plus seven is fourteen (7*2=14), plus seven is 21 (7*3=21)

18.75% per annum compounded monthly would result in 18.75/12 = 1.5625% per month

So we have  459(1.15625)⁶ = $6679.33 owing

After some consideration, I have to say, I'm not sure WHAT  the answer to this question might be.

For instance, in this case, if something were sped up by 50%....50% of 15 = 7.5 sec. So now, it should take only half as long. But, by the same token, if it was sped up by 100%, it wouldn't take any time at all, since 100% of 15 is 15.....!!!

I guess that I'm unsure if something can be sped up by 200%....???

Any constructive criticism is welcome, here...!!!

In light of this, maybe Melody's answer of 5 seconds represents about a 67% increase???....(since 67% of 15 = 10.....???)

a)

AK=h     (right isosceles triangle AKH)

$$\\tan30=\frac{HK}{KD}\\\\
\frac{1}{\sqrt3}=\frac{h}{KD}\\\\
KD=\sqrt3\;h\\\\\\
AH^2+KD^2=AD^2\\
h^2+(\sqrt3\;h)^2=AD^2\\
AD^2=h^2+3h^2\\
AD^2=4h^2\\
AD=2h$$

b)

It takes 20 min to walk AD

15min to walk DE

and 15min to walk AE

so triangle ADE is isoceles and the ratio of the sides is 4:3:3

AD is 2h  so   DE=AE= (2h/4)*3 = 1.5h

Usine cosine rule 

$$\\cos(E)=\frac{3^2+3^2-4^2}{2*3*3}\\\\
cos(E)=\frac{9+9-16}{18}\\\\
cos(E)=\frac{1}{9}\\$$

$$\\$Area of \triangle ADE = $0.5*AE*ED*sinE$$

$$\\6272\sqrt5=0.5*1.5h*1.5h*sin(cos^{-1}\;\frac{1}{9}\;)\\\\
6272\sqrt5=0.5*1.5h*1.5h*\frac{\sqrt{80}}{9}\\\\
6272\sqrt5=\frac{1}{2}*\frac{3h}{2}*\frac{3h}{2}*\frac{4\sqrt{5}}
{9}\\\\
6272\sqrt5=\frac{9h^2}{8}*\frac{4\sqrt{5}}{9}\\\\
6272=\frac{h^2}{2}\\\\
h^2=12544\\\\
h=112$$

I simultaneously worked the problem and wrote the code so there could easily be error.

h did turn out to be a 'nice' number, so maybe it is right :)))

Feel free to ask questions 

I just realized that it is not finished yet.

I think the minumum angle of elevation TO H   is from E because that point is furthest away

I need to find distance EK

$$\\KD=\sqrt3*h\\
AE=ED=1.5h$$

This is not just falling out and I don't have more time now.  I shall return if another doesn't answer first. 

0.1666666666666667

Is the calculator is able to lay out the Newton polynomial ?

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Yes matHunter! It's good translated question.

with a car dealership price of $19,192, excluding sales tax the sales tax rate is 8.375%

a. what growth factor is associated with sales tax rate?  1+0.08375 = 1.08375   (or  108.375%)

b.total price of the car is $?      $19,192 * 1.08375 = and you can finish it :)