This has taken a while. I think this works:
Construct the perpendicular bisectors of AD, BD and DC. The circumcentres are at the intersections so I've marked these as E and F.
H is the midpoint of BD, G of DC. I've also drawn JK parallel to BC through A
As FD = ED, the line AD bisects that isosceles triangle, and GHKJ is a rectangle (shaded yellow).
Area ABC = half base x height = 0.5 x BC x HK. But as H and G are midpoints 0.5 x BC = GH, so the shaded rectangle is the same area as ABC.
We know one side (GH) and can calculate the other like this.
A perpendicular from point A to BC will meet BC at L (not shown). Triangle ALD is right angled with AD = 13, and LD = 7-2 = 5. So you can use Pythagoras to calculate AL and hence complete the question.
Triangle ABC is inscribed in equilateral triangle PQR, as shown. If PC = 3, BP = CQ = 2, and
, then compute AQ.
I think I need more information as there is no diagram. Please say which side each of A, B and C is. eg. B is on PR.
To get those squared terms there must be two factors each with an x term and a y term. To get minus y squared one must have a minus sign. So it would take the form:
(x + y + a)(x - y +b) for some numbers a and b. That also accounts for no xy term.
So that makes
x^2 - xy + bx + yx - y^2 + by + ax - ay + ab.
From the x coefficients b + a = -6 and from the y coefficients b - a = 4
That's enough to fix a and b. Adding gives 2b = -2 so b = -1 and so a = -5.
This gives ab = 5 which is just what we wanted, so here's the factorisation:
(x + y -5)(x - y - 1)
Many people found the film incomprehensible and I would too, except I am a keen Arthur C Clarke fan, and I had read the story "The Sentinel", on which the film is based. I can recommend this story and it makes sense. According to Wiki it was published in a collection of short stories under the same name.
We are several million miles apart. This is the HAL that I was referring to:
But hey! Never mind. I spent some educational moments reading about quotes from the bible and misinterpretations of those. Conspiracy theorists may be able to find a connection, given the film title and subject matter.
Hopefully what follows is a useful way to do this problem. It's still a little long but loads of useful properties drop out during the proof. One of the problems when the diagram is complicated and accurate is it becomes very easy to assume something that needs to be proved. I think I've ironed out any such bugs but please let me know if you spot any **.
In this diagram AB is parallel to CD as given. E and F are the circle centres. G is the midpoint of AB.
GA = GP = GB by the equal length tangents theorem. So a circle centred on G will go through A, B and P.
It will also cut the line CD again whilst the original circles are unequal in size. Let H be the point where the circle cuts CD. Continue HG back across the circle to point L.
Let angle GAH be x and angle GBH be y.
As H is subtended by a diameter AHB = 90, and as BGH and AGH are isosceles, GHB = y and GHA = x, and x + y = 90.
Let BF cut CD at J.
As ABF is 90 and AB is parallel to CD, BJC = 90, and so PJ = JD.
Let AE (extended in my diagram) cut CD at K.
Then, in the same way, AKD = 90 And CK = KP.
So ABJK is a rectangle.
CD = CK + KP + PJ + JD = 2(KP + PJ) = 2KJ = 2AB.
I have just spotted a **. I have assumed that DB produced and CA produced meet at L. I shall have to prove that.
Everything falls into place once I've done this so there will be a (hopefully short) intermission while I get this step sorted.
OK. I think I've got it.
Angle LGB = AGH (vertically opposite) and triangle GBL is isosceles so LBG = x.
At point B we have angles of x, y, and x so JBD = y and BDJ = x. So DB extended goes through L. Similarly CA extended so L is the point where DB and CA cross.
Lots of results follow:
(1) triangle LAB is similar to LCD and half its size.
(2)LBHA is a rectangle.
(3)ABDH has a pair of parallels and opposite angles equal so it is a parallelogram.
(3)Similarly, ABHC is a parallelogram.
(4)So it follows that H is the midpoint of CD.
(5)GH is half AB.
I hadn't spotted that. Once I've done the problem my way, I can easily show the properties you want, but that doesn't help if you want to develop a fresh solution. I've had a longer and more thorough search for an earlier version of this problem and found nothing. So I think I must have dreamed that I've seen it before.
I feel there must be an easier way to do this (it's only CompuHigh after all ) so I'm going to start again with a fresh diagram. Wish me luck.
Welcome to the forum.
A new zero will be added when a new factor of 2 and 5 occurs, as 2x5 = 10. As every other number has 2 as a factor there won't be a shortage of them, so it's down to when does a factor of 5 occur. The first time is 5! = 24 x 5 = 120. Then 10! (=3628800) will add another and so on.
Writing a factorial as a product of primes should make this clear:
10! = (5x2) x (3x3) x (2x2x2) x 7 x (3x2) x 5 x (2x2) x 3 x 2 x 1
I had rounded it off. 22/7 would make a slightly different answer, but it would come to the same when rounded. If you think about the problem in terms of accuracy, it makes sense to round things. The height and radius are given as integers. As it is impossible to measure something with absolute accuracy you should take into account what difference a small change like 18.1 rather than 18 would make. And paint coverage introduces a huge potential inaccuracy. The exact coverage will depend on how much the painter has on his brush; that figure will always be only an estimate.
In GCSE (UK exam) it says in the syllabus that candidates should round off appropriately and markers are given a range of answers to take account of this. Just because your calculator can work to (say) 10 figures, doesn't mean you have to declare them all when giving an answer.
The s and t method is much the same as what I suggested, so I'm happy with that.
It isn't factorable and that solution is right:
You can get x from those two values but it's tricky. How about doing it like this:
times throughout by x^2
Then you can use the quadratic formula on that for immediate solutions.
The formula for the curved surface of a cylinder (imagine it 'unfolded' to make a rectangle) is
and for the two ends, two lots of
So add up the total surface area and times by the cost per square not cube centimetre.
I get 407 for Q2.
Q3. If the sphere exactly fits inside the cylinder then they have the same diameter (and radius) and the height of the cylinder must also be the diameter of the sphere.
Q1. As you know that x = -3 or 2/3 you can say that (x+3) and (3x-2) are factors. Multiply them and you'll get a quadratic. You can pick out values for a, b and c from this.
[pedantic note (ignore if you wish): So we end up with numbers to fit ax^2 + bx + c = 0. But those solutions will also fit akx^2 + bkx + ck = 0 for any integer k. So 2a + b - c could take the value k(2a + b - c). In other words any number in a certain times table is a possible answer.
Q2. I would call the solutions alpha and beta. So
So you can write
Make alpha the subject of one of these and substitute into the other. You get a quadratic which will tell you b and c. And you can then solve it. (note: If instead, you seek to eliminate beta you'll end up with the same quadratic so it doesn't matter which way you do it.)
Q3. It takes ages to format this nicely for a post so I hope you know roughly what this method involves.
step 1. To get -x^4 you need to multiply the divisor by -2x^3. Subtracting gives -4x^3.
step 2. To get that you need to multiply the divisor by +4x^2. Subtracting gives 11x^2.
I made the final remainder 52. Hope that is enough.
The method is nicely demonstrated here: http://www.mathsisfun.com/algebra/polyn … -long.html
No, no, no. Not bonehead. You got a special award (gold star) earlier. It just seemed to me that all those curves made it impossible, so there ought to be a way to get rid of them. Did you like the diagram ? Confucius, he say, "A picture is worth ten thousand words."
That's twice his name has come up in the last half hour.
Is there a method ? (-3,2) is pointing me towards the equation of a straight line http://www.mathsisfun.com/algebra/syste … tions.html
The given line goes through that point and so will any other equation for which this is the solution.
Sometimes my brain is so sure I've got it sorted I forget to pay attention to what I'm saying. I could see a calculation was possible there; just got the wrong one. I ought to have learnt by now but I haven't. In fact I seem to be getting worse. Just keep asking until I start making sense. .
Welcome to the forum.
What do you mean by inversion?
In 'real' encryption the value of n is a lot bigger. Graph plotting may not be sufficiently accurate to determine the factors. Actually, I hope this is the case (sorry) or it blows a hole in encryption and I'm quite keen that we can continue using it for a while yet.