What is the 3,4,5 rule in finding a 45 degree angel?
I'm guessing what your looking for is the old carpentry trick for squaring up perpendicular lines. To use the 3, 4, 5, rule you measure and mark 3 feet from the intersection on one line, and 4 feet on the other line. The measurement diagonally from mark to mark will be exactly 5 feet only if the angle is exactly 90 degrees. BTW This trick can be made more versatile by applying a little more algebra to the Pythagorean theorem. nA^2 + nB^2 = nC^2 is an equivalent equation and the implication is that you can multiply all sides of a 3, 4, 5, triangle by an equal factor. The advantage for this is as follows. Say you are checking the walls of a building 20' x 25' to see if they are square. You then find a factor you can multiply 3 and 4 by that's as large as possible without exceeding the wall length. I this case a factor of 6 works best. So then you measure 18' (3'x6) down the 20' wall and 24' (4'x6) down the 25' wall and then measuring diagonally you will get a measurement of exactly 30' (5'x6) if the walls are at a perfect 90 degree angle. You can also use this trick for objects smaller than 3' x 4' by using 3, 4, and 5 inches instead of feet. For example you want to square a picture frame 22" x 32". The greatest factor in this case is 7. Thus you make a mark at 21" (3" x 7) and at 28" (4" x 7) and the measurement from mark to mark diagonally will be 35" when square. On Square or rectangluar objects it's actually easier to check the corners for squareness by making sure that the length from one corner to the diagonally opposite corner equals the length between the remaining two corners. One last trick is if your using the same dimensions for repeated checks you may find it convienient and more accurate to cut a stick the length of the hypotenuse instead of trying to hold and read a kinked measuring tape.
Let A & B be two sets containing four & two elements. Then, what will be the number of subsets of set A×B each having at least three elements?
Let Set A = {1,2,3,4} & Set B = { a,b}=> A x B contains 4*2 = 8 elements ( or ordered pairs),Like A x B = { (1,a),(1,b),(2,a),(2,b),(3,a),(3,b),(4,a)(4,b) }So, A x B contains 8 ordered pairs.Now we need to find the subsets of A x B , in which at least 3 elements ie 3 ordered pairs should be there. That means it can have 8 ordered pairs or 7 ordered pairs or 6pairs or 5pairs or 4pairs or 3 pairs. It can not go below 3 pairs as question is we need subsets with at least 3 pairs.So we start making the subsets with 8 pairs.●With 8 ordered pairs…. we have just 1 subset , which is {(1,a),(1,b),(2,a,),(2,b),(3,a),(3,b),(4,a),(4,b)} (as such , every set is its own subset)●similarly with 7 ordered pairs : the number of ways of choosing 7 pairs from the set of 8 pairs will be= 8C7 = 8!/(1! * 7!) = 8 subsets● Now similarly with 6 pairs : we get 8C6= 8!/(2! * 6!) = 28 subsets● With 5 pairs : we get 8C5 = 8!/(3!*5!) = 56 subsets● With 4 pairs : we get 8C4 = 8!/(4!*4!) = 70 subsets●Now with 3 pairs: we get 8C3 = 8!/(5!*3!) = 56 subsetsNow, by adding all the above subsets , we get..1 + 8 + 28 + 56 + 70 + 56 = 219 subsets . . . . Ans
Let a and b belong to a ring R and let m be an integer. Prove m(ab)=(ma)b=a(mb)?
[I'm afraid I have no way of knowing what "rules of Mult #2" might mean]. The most straightforward proof method is by induction. Firstly the identity holds for m = 0: 0 = 0.(ab) = (0.a)b = a(0.b). Now suppose the identity holds for m = n; then we have n.(ab) = (n.a)b = a(n.b) ⇒ n.(ab) + ab = (n.a)b + ab = a(n.b) + ab ⇒ n.(ab) + 1.(ab) = (n.a)b + (1.a)b = a(n.b) + a(1.b) ⇒ (n + 1).(ab) = (n.a + 1.a)b = a(n.b + 1.b) ⇒ (n + 1).(ab) = ((n + 1).a)b = a((n + 1).b). The case where m is negative follows using a similar argument (or by applying the identity a(-b) = (-a)b = -(ab)).
How many functions f(x) satisfy this relation?
The following solution is over the real numbers. If a and b are really constants, there are very many solutions since f can behave arbitrarily away from a, b, and a+b. If a and b are allowed to be arbitrary, and we assume that f is twice differentiable everywhere, we have the identity f'(a + b) = ∂(f(a+b)) / ∂a = ∂(f(a+b)) / ∂b but ∂(f(a+b)) / ∂a = f'(a) f'(b) + f''(a) f(b), ∂(f(a+b)) / ∂b = f(a) f''(b) + f'(a) f'(b) so f''(a) f(b) = f(a) f''(b). Unless f is zero everywhere, we can find some b with f(b) ≠ 0, and we then have the differential equation f''(x) = K f(x), where K := f''(b) / f(b). We now have one of the following three cases: (1) f(x) = C exp(x √K) + D exp(-x √K), for some constants C, D, if K > 0, (2) f(x) = C x + D, if K = 0, or (3) f(x) = C sin(x √(-K)) + D cos(x √(-K)), for some constants C, D, if K < 0. We plug f(x) back into f(a + b) = f(a)f'(b) + f'(a)f(b) and solve for C and D. Omitting some algebraic steps here: In case (1), we get C = 0 or 1/(2√K) and D = 0 or -1/(2√K). In case (2), we get C = 0 or 1 and D = 0. In case (3), we get D = 0 and C = 0 or 1/√-K. This exhausts all possible solutions. So, if the given relation is an identity in a and b, and f is twice differentiable everywhere, the only solutions are f(x) = 0, f(x) = x, f(x) = exp(Lx) / (2L), L a nonzero constant, f(x) = sinh(Lx) / L, L a positive constant, f(x) = sin(Lx) / L, L a positive constant.
If log2 = a and log 3 = b , then express log 25/8 in terms of a and b?
Find the cosine of the angle between the planes 1x - 4y - 5z = 1 and the plane 5x - 2y +3z = 4?
The normal directions to the planes can be read from the equations. The plane ax + by + cz = d has normal direction (a,b,c). The angle between the planes is the same as the angle between their normal vectors. The normals are n1= (1,-4,-5) and n2 = (5, -2, 3). The cosine between them is given in terms of the dot product. cosΘ = n1∙n2/(||n1|| ||n2||). n1∙n2 = 5 + 8 - 15 = -2 ||n1|| = √(1 + 16 + 25) = √(42) ||n2|| = √(25 + 4 + 9) = √(38). So cosΘ = -2/√(38*42) = -1/√(399).
Chemistry question?
1) In what region of the electromagnetic spectrum would you look for the spectral line resulting from the electronic transition from the fifth to the tenth electronics level in the hydrogen atom. (Rh=1.10 x 10^-5 cm^-1) 2) Use the Rydberg equation to calculate the frequency of a line in the hydrogen spectrum corresponding to a transition from n=5 to n=4. Identify the spectral region or color which corresponds to this frequency.
If A and B are disjointed sets, how can you find n (A union B)?
Two sets are said to be disjoint if they have no element in common. Equivalently, disjoint sets are sets whose intersection is the empty set. For example, {1, 2, 3} and {4, 5, 6} are disjoint sets,( as they have no element in common) while {1, 2, 3} and {3, 4, 5} are not.(as the element 3 is in common).The union of two sets A and B is the set of elements which are in A, in B, or in both A and B. In figures;First consider the case where the sets A and B are disjoint.In that case,The number of elements in the union (A∪B) is simply the sum of the number of elements in A and the number of elements in B: |A ∪ B| = |A| + |B|. [ |A|→no of elements in A and other notations mean similar].But if A and B overlap, then the latter formula does not hold because, we are counting the elements in the intersection (A ∩ B) twice. Compensating for that leads to the given formula: |A ��� B| = |A| + |B| − |A ∩ B|.[ Note : n(A U B) is also denoted as |A U B| ]In above figure of Disjoint sets;Elements in (AUB)=elements in (A)+ elements in (B).In above example, union of disjoint sets is;Element set in A + Element set in B={1,2,3,5,7,9}.
I need help with this math problem. Can you help me?
A graphic artist needs to construct a design that uses a rectangle whose length is 5 cm longer than its width x. (a) Construct a model that gives the perimeter P(x) of the rectangle.