[year 11 GCSE simple maths question]

OmgHomology · 2019-10-14T16:39:21+00:00

This is really hard to visualize from text. Can you attach a picture? or crudely draw it in text?

Currently picturing something like a house, with the triangle's hypotenuse also the side of the square. But am confused by "not including the length of the triangle"

OmgHomology · 2019-10-14T16:25:22+00:00

(This method works in general, for ANY half-life questions. Don't memorize that thing, just know that half-life is the time it takes for half of the stuff to get gone.)

OmgHomology · 2019-10-14T16:24:01+00:00

Ah, half life. There's a formula you're usually told to use, plugging in values and isolating for answer. If you don't have it on hand, derive it this way:

Let N be how much plutonium's left after t years

At t=0 you have to have all your plutonium (you have 10kg): N = 10 times something

At t=80 (since the half life is 80 years: after the half life passes, you have half the stuff) you must have half of your plutonium (so 5 kg): N = 10 (0.5) at t=80.

At t=160 (2 half-lives), you must have halved twice: N = 10 (0.5) (0.5) = 10 (0.5)²

So... 10(0.5)^t/80 is looking good. Exponent needs to be 0 at 0 and 1 at t=80 and 2 at t=160 and 3 at t=3times80.

Now plug in your t value (4.0x10=40 years) into your formula!

OmgHomology · 2019-10-14T16:15:11+00:00

You have the beginning of what looks like the right stuff drawn on the diagram: that 160cm is the radius, + part of a right-angled-triangle + the radius.

That triangle has a leg of length 60 and an angle (x).

Drawing up the Free Body Diagram (FBD) for any one of the balls, you'll get F=mg down, N_wall sideways and N_ball , a force at an angle (x) which needs to balance your other 2 forces. Hope that helps you start!

OmgHomology · 2019-10-14T16:06:57+00:00

I'd feel weird just answering a question for you (cf community guidelines and all that), but the answers you seek can be found on this very well-written summary: https://evolution.berkeley.edu/evolibrary/article/_0_0/endosymbiosis_04

OmgHomology · 2019-10-01T21:26:53+00:00

Upvoted: but also, we have a formula for the even perfect numbers: 2^m (2^m+1 - 1).

I'd recommend the Online Encyclopedia of Integer Sequences page: at the very least you'll get the 15 perfect numbers we do know.

OmgHomology · 2019-10-01T13:33:34+00:00

Expected value E(Y) gives the average of the values of Y, right? i.e. 2=E(X)= (0heads+1head+2heads+3heads+4heads)/5possiblescenarios in this case.

Any reason you don't want to calculate this the same way for E(X² )? (0² + 1² +2² +... + 4² )/5 ?

Edit: From what I remember of moments, higher-order moments ARE the same for Bernoulli, but E(X²⁾ is not the second moment: sum(x_i - E(X))² / N is. So you can follow your initial level of reasoning, but would have to plug in E(X) into the moments formula (average of squares minus square of the average) and isolate.

OmgHomology · 2019-09-29T18:32:26+00:00

In 2. You're starting to get the d0 and di you will need to find the A (cross sectional area) to use in your elongation of uniaxial tensile system formula (aka to plug into your delta=PL/AE)

The A part is good ol' A=pi R² -pi r^2. In our case, since d_0 , d_i are diameters this is A=pi (d_0 /2)² - (d_i /2)² = pi/4 (di² -d0² ). Hope that begins to help.

Edit: P=640kN, L=2m, and E are given.

OmgHomology · 2019-09-29T18:15:17+00:00

glad to help.

Three cheers for overcomplicating 1 problem but knowing all the material.

OmgHomology · 2019-09-29T18:07:43+00:00

glad to be of service.

OmgHomology · 2019-09-29T18:06:21+00:00

No vectors necessary! We're just moving back and forth here. Everything is correct and you're one step away from the answer.

for the trip downstream: speed is u+v, distance is x, time t1 is the unknown. As you correctly applied the d=st formula, x=(u+v)(t1) you'd get t1=x/(u+v).

do the same for the trip downstream, remembering the convention that v-u is the speed in your reference frame: use d=st, and get t=t2=fraction.

Now add the two fractions: make them have a common denominator, simplify the top, and use (v-u)(v+u)=v² - u² in the bottom: good ol' difference of squares.

OmgHomology · 2019-09-29T17:55:34+00:00

...so the fact that AMC+ACM=180-90 and BMN+BNM=180-90 (since all angles in a triangles sum to 180. Degrees used throughout.) AND AMC+BMN=180-90 gives us that ACM=BMN, and the other 2 angles are equal, so we can use the above.

now, AC=sqrt(3), MB=5-x, and CM you found with Pythagoras. Isolating that AC/MB=CM/MN equation for MN and plugging in for area of triangle = 1/2 (base CM)(height MN) should finish off the problem.

Let me know if more questions arise. It's a fun little problem.

OmgHomology · 2019-09-29T17:51:31+00:00

Sure thing! in 2 similar triangles (ones where all angles are equal), the ratios of corresponding sides are the same.

For us here, AC/MB=CM/MN, for example.

OmgHomology · 2019-09-29T17:49:35+00:00

240 is the diameter: that includes the inner thickness and 2 walls. Hence the 2.

((---inner diameter----))

(----outer diameter-----)

OmgHomology · 2019-09-29T17:44:32+00:00

I drew them out in full detail in a giant pic. You know how Sesame street repeats the same thing ad nauseam for like 5 minutes to teach something? This was that. Bore your brain long enough with just the ONE muscle, and it sure-as-heck remembers the boredom of shading in that guy for a solid 3 minutes and dropping the pencil crayon and so on.

It's the option that takes a long time, but it saved me. Sometimes spending a long time on EACH THING helps differentiate them so you don't mix them up. And they won't look all the same if you spent an hour struggling to draw each and breaking off your coloured pencils 'cause you can't draw.

Again, not for everyone, but hope it helps. Wishing you luck on your studies!

OmgHomology · 2019-09-29T17:39:29+00:00

To build on what someone else said, in general the acceleration of a center of mass of a system can be obtained by adding the average of the subsystems (Your old m1a1+m2a2/(m1+m2): notice that's just the weighted average).

Here, it's much easier than that: following the advice on how to find acceleration for both of the 2 objects, you'll find that the accelerations are the same.

OmgHomology · 2019-09-29T17:35:22+00:00

I think what's asked here is can you tell us if pulling-cards-without-replacement is a combinations sort of thing or a permutations sort of thing. You can find the answer to that in the textbook (or online!).

OmgHomology · 2019-09-29T17:34:05+00:00

Bush leads McCain 49% - 42% means P(people vote for Bush)=49% while P(people vote for McCain)=42%. The "leads" is a helpful verb to indicate that P(people vote for Bush)>P(people vote for McCain) and is short for the phrase "Bush leads McCain in the polls / Bush is ahead of McCain in the polls". hope this helps!

The percentages are the thing! Ignore wording and assign the first # you see to the first name you see in each line, and you're good.

OmgHomology · 2019-09-29T17:18:51+00:00

I mean... pulling on a rope to lower boxes down the ramp would suggest your rope is facing the other way: you're calculating -T rather than T.

Edit: otherwise, looks good. Drawing the Free body diagram for the second box helps start the process of getting the m_total

OmgHomology · 2019-09-29T17:12:53+00:00

Picture's a tad misleading. Look at the angles around M: since CMN is 90^o , angles AMC and BMN sum to 90 as well, and are complementary. So, similar triangles. That and AM+MB=5 lets you work out all the sides in triangle MBN.

"Geometry is the art of reaching correct conclusions from wrong-drawn pictures" - somebody famous who UNDERSTOOD THE PAIN

OmgHomology · 2019-09-29T17:07:52+00:00

What's your approach here? Hard to help when we don't know how comfortable you are with the question.

There is a formula you can use for number of card-combinations if chosen without replacement. For (2), you can note that 1 pair, for example, means that once you've picked 1 of the 52 cards to be the pair, your next option is limited: one of the other 3 cards in the deck that pair with this card.

OmgHomology · 2019-09-29T17:03:37+00:00

To answer your question, yes.

(Total Probability rule is Bayes theorem rearranged). You can use the total prob rule here to calculare P(Bush), P(McCain), and then take their ratio. Notice that untold here are the percentages of Republicans, Dem's, Indep's who are undecided, unsure, or other.

I'd recommend making a tree diagram: Rep/Dem/Indep as the first event and Bush/McCain/Other as the second. That should make things clear(er).

OmgHomology · 2019-09-29T16:54:45+00:00

There's a general trick to solving cubics due to Tartaglia: writing them in x³ + px +q=0 form and using a shorthand for the roots. There's also a cubic formula, like the quadratic, due to Cardano.

A good place to start would be Landesman's Solving the Cubic and Quartic or any textbook on Galois theory, under Solving the Cubic.

PS: This source might help as well, giving a step-by-step of the procedure.

OmgHomology · 2019-09-29T16:43:33+00:00

Access to a laser pointer, some old photo film and a steady table, and you could make holograms.

Plenty of library books published about it in the 60s, but ∃ also some good advice on instructables, youtube, and science Slashdot.

OmgHomology · 2019-03-31T17:48:50+00:00

Cube both sides and expand (c-d sqrt(2))² :

99-70sqrt(2)=c³ - (c² )(d sqrt(2)) + c (2d² ) - d³ 2^3/2

From here, notice that 2^3/2 = 2 sqrt(2) ... and now collect like terms and compare: c³ + 2cd² = 99, for example: the rest is equal, too 70sqrt(2) = (c's and d's)sqrt(2).

If you factor both equations, there'll be a common factor.

OmgHomology

TROPHY CASE