(tg^2x-sin^2x) / (ctg^2x-cos^2x) = tg^6x Demonstration pliz.. Help me!!!?
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The product of two numbers is 65. What is the sum of the two numbers?
The product of two numbers is 65. What is the sum of the two numbers?Before we do any hard math computation let us examine the question. The question is simply asking “add the factors of 65”. There are two that comes to mind easily namely (numbers in bold print):65 x 1 = 6513 x 5 = 65Now add the factors above and you arrive at two possible answers.65 + 1 = 6613 + 5 = 18Answer: The sum of the two numbers is either66 or 18
The product of two numbers is 15. If one number is 5, what is the other number?
We’re doing someone’s homework again.You’ve been told that the product of two numbers is 5. Product refers to the outcome of a multiplication, so you know that something multiplied by 5 gives you 15.Let’s us an inverse: if we’ve multiplied 5 by x to get 15, dividing 15 by 5 will give us x.5x = 1515 / 5 = 3x =3
What is the maximum product of two numbers whose sum is −16? What numbers yield this product?
The Product is this: [math]P(a, b) = a*b[/math]That’s the first-order condition.And you know: [math]a + b = -16[/math]That’s the second-order condition.Now You can rearrange the second equation:[math]a + b = -16[/math][math]b = -16 - a[/math]That equation can subsitute the b in the first-order condition:Therefore:[math]P(a, b) = a*b[/math][math]P(a) = a*(-16 - a)[/math]Now, you have two possibilities:Complete the square and read off the vertex of the parabolaMultiply out and derive the function to get the extreme valueLet’s do 1 at first:[math]P(a) = a*(-16 - a)[/math][math]P(a) = -16*a - a*a[/math][math]P(a) = -a*2 + -16a[/math][math]P(a) = -(a^2 + 16a)[/math][math]P(a) = -(a^2 + 16a + \dfrac{16}{2}^2 - \dfrac{16}{2}^2)[/math][math]P(a) = -(a^2 + 16a + 8^2 - 8^2)[/math][math]P(a) = -((a + 8)^2 - 8^2)[/math][math]P(a) = -(a + 8)^2 + 8^2[/math][math]P(a) = -(a + 8)^2 + 64[/math]That gives us: [math]S(-8 | 64)[/math]Plug in in the second equation:[math]b = -16 - a = -16 - (-8) = -16 + 8 = -8[/math]So, [math]a = -8[/math] and [math]b = -8[/math].The product is 64 (the y-value of the vertex).Let ‘s try out the second way as well:[math]P(a) = a*(-16 - a)[/math][math]P(a) = -a^2 - 16a[/math]Derive:[math]P'(x) = -2a - 16[/math][math]P'(x) = 0[/math] to get the turning point.[math]0 = -2a - 16[/math][math]16 = -2a[/math][math]a = \dfrac{16}{-2} = -8[/math]And: [math]b = -16 - a = -16 - (-8) = -8[/math][math]a = -8[/math], [math]b = -8[/math]So, you get the same result in both cases.Hope this helps.
How do I write a C program to calculate the sum of two numbers?
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What is the LCM of two co-prime numbers, x and y?
Here x and y are co-prime which means they do not have any common factor other than 1. Co-prime do no necessary have to be prime numbers. They are just prime with respect to each other. The product of co-prime numbers is the LCM. Here, it is xy.For example, let x= 9 and y=5. They are co-primes. Their LCM is xy= 9×5= 45.Hope it helps!
Can multiplication of two numbers equal zero if neither of them is zero?
In the context of a field, such as ordinary addition and multiplication on the rational numbers or on the real numbers or on the complex numbers, the product of two numbers is 0 if and only if at least one of the numbers is 0.If one goes to a different type of algebraic structure with addition and multiplication operating on a set of values that does not constitute a field, then all bets are off. For example in the ring of modular addition and multiplication modulo 6, we have 2 × 3 = 0, because to evaluate the result, you first multiply 2 × 3 to get 6 and then divide that result by 6 to get an integer quotient and a remainder ranging from 0 to 5. Dividing 6 by 6 yields of quotient of 1 with a remainder of 0, so 0 is the result—and, yes, the product of 0 with anything in the set is 0. In this case 2 and 3 are called zero-divisors (with zero divisors being any nonzero numbers in the set that can be multiplied by some nonzero number in the set and end up with a product of 0—fields have no zero divisors).
What is the proof, if it is true, that the product of two positive integers greater than 1 is greater than or equal to their sum?
Long answer coming up.We can prove this problem by mathematical contradiction. Now, given 2 numbers x and y, three possibilities exist.1. xy = x + y2. xy > x + y3. xy < x + yIt is enough if we prove that the third statement is a contradictory one.Here, we are faced with 3 cases. 1. x = y2. x > y3. x < y1.Given that x = y. :By proposition, xy < x + y.Implies , x^2 < 2x. But this is a contradiction. Since for all x>1, x^2 > 2x.2.Given that x>y:Let y = x + c. By proposition, xy < x+yImplies, x^2 + xc < 2x + cImplies x^2 - 2x < c ( 1-x )We know that x^2 - 2x >1. Thus, c(1-x) > 1. But since 1-x < 0 , c(1-x) < 0. Clearly a contradiction. [ we know that 1-x < 0 since x is a positive number greater than 1 which is out basic assumption ]3. Given x < y. This can proved along similar lines.So, what we have done now , is proven that xy < x + y is not possible. Thus, only the first two cases remain.QED.Is this solution acceptable?
Prove that there are real numbers a and b such that sqrt(a+b) = sqrt(a) + sqrt(b)?
Because I think there needs to be an answer here that isn't obnoxiously condescending, here it goes. Suppose that there are some numbers a and b such that √(a+b)=√a + √b. In order to simplify the equality, and figure out which values of a and b would work, we should get rid of as many square roots as possible. √(a+b) = √a + √b... square both sides a + b = a + 2√(ab) +b... subtract a + b 0 = 2√(ab).. we could stop here, or divide by 2 and square to get 0 = ab Anyway, if the product of 2 numbers is zero, one of them has to be zero. Since each equality on the list implies the one above or below it, we know that the original equality holds if and only if one of a and b are equal to zero.
What is the HCF and LCM of fractions?
Let’s take two fractions (a/b) and (c/d).To find LCM and HCF of (a/b) and (c/d) the generalized formula will be:H.C.F = H.C.F of numerators / L.C.M of denominatorsL.C.M = L.C.M of numerators / H.C.F of denominatorsNow L.C.M of two numbers is the smallest number (not zero) that is a multiple of both. Eg: L.C.M of 12,15:The multiples of 12 are : 12, 24, 36, 48, 60, 72, 84, ....The multiples of 15 are : 15, 30, 45, 60, 75, 90, ....60 is a common multiple (a multiple of both 12 and 15), and there are no lower common multiples.Therefore, the lowest common multiple(L.C.M) of 12 and 15 is 60.H.C.F of two numbers is the largest whole number which is a factor of both. Eg: H.C.F of 12,15:The factors of 12 are : 1, 2, 3, 4, 6, 12.The factors of 15 are : 1, 3, 5, 15.1 and 3 are the only common factors (numbers which are factors of both 12 and 15).Therefore, the highest common factor of 12 and 15 is 3.Now,to find out the LCM and HCF of fractions .Let’s take example of (4/5) & (3/7):LCM=lcm of (4,3)/hcf of (5,7) = 12/1 = 12 (Since 12 is the least number that comes in table of both 4,3 and 1 is the greatest number that can divide both 5,7).HCF=hcf of (4,3)/lcm of (5,7) = 1/35 (Since 1 is the greatest number that can divide both 4,3 and 35 is the smallest number that comes in table of both 5,7).I hope you get it!Shukriya!