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Find HCF on 285 and 1294 whose remainders are9 and 7

find HCF  of  285and1249 whose remainders are 9 and 7

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Find HCF on 285 and 1294 whose remainders are9 and 7

me dont understand the quesshun "HCF" normally meen hiest konnom fakter, but 285 & 1294 dont hav NE kommon fakters Then yu talk bout "remaenders".  Fakters defined==number that divide intu start number with leftover=0.

find HCF of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3

find HCF and LCM of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   LCM=(x-1)(x+1)(x+3)(3x-1)(x+2)(2x-1) contains a combination of the factors of both numbers. By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3

find a geometric progression whose sum is 5/2. Is it possible to find another one?

In an infinite GP converging to a finite limit, S=a/(1-r) because Sn=a(r^n-1)/(r-1)=a(1-r^n)/(1-r). When r is small (<1) and n is large r^n approaches zero and is zero when n=infinity. Therefore, 5/2=a/(1-r) so 5(1-r)=2a. This shows a relationship between a and r for 0 Read More: ...

Find the volume of the solid whose base is the region bounded by y=x^7, y=1,

The radius of each semicircle is (1-y)/2 which equals (1-x^7)/2. The area of such a semicircle is π((1-y)/2)^2/2=π(1-y)^2/8. If the semicircle has a thickness dx the volume of the semicircular section is π(1-y)^2dx/8. When y=1, x=1, and (0,0) is the starting point. The volume of the first semicircular slice is when y=0, making the volume πdx/8, and the volume of the last semicircular slice is 0. So (π/8)∫((1-x^7)^2dx) for 0≤x≤1 gives us the volume. Expanding this we have: (π/8)∫((1-2x^7+x^14)dx)=(π/8)[x-x^8/4+x^15/15] for 0≤x≤1. This gives us (π/8)(1-1/4+1/15)=(π/480)(60-15+4)=49π/480=0.3207 approx.

i need the answer for these questions

Part 1 Newton’s Method for Vector-Valued Functions Our system of equations is, f1(x,y,z) = 0 f2(x,y,z) = 0 f3(x,y,z) = 0 with, f1(x,y,z) = xyz – x^2 + y^2 – 1.34 f2(x,y,z) = xy –z^2 – 0.09 f3(x,y,z) = e^x + e^y + z – 0.41 we can think of (x,y,z) as a vector x and (f1,f2,f3) as a vector-valued function f. With this notation, we can write the system of equations as, f(x) = 0 i.e. we wish to find a vector x that makes the vector function f equal to the zero vector. Linear Approximation for Vector Functions In the single variable case, Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus, the following is the linear approximation of f at x0, for vectors and vector-valued functions: f(x) ≈ f(x0) + Df(x0)(x − x0). Here Df(x0) is a 3 × 3 matrix whose entries are the various partial derivative of the components of f. Speciﬁcally,     ∂f1/ ∂x (x0) ∂f1/ ∂y (x0) ∂f1/ ∂z (x0) Df(x0) = ∂f2/ ∂x (x0) ∂f2/ ∂y (x0) ∂f2/ ∂z (x0)     ∂f3/ ∂x (x0) ∂f3/ ∂y (x0) ∂f3/ ∂z (x0)   Newton’s Method We wish to find x that makes f equal to the zero vector, so let’s choose x = x1 so that f(x0) + Df(x0)(x1 − x0) = f(x) =  0. Since Df(x0)) is a square matrix, we can solve this equation by x1 = x0 − (Df(x0))^(−1)f(x0), provided that the inverse exists. The formula is the vector equivalent of the Newton’s method formula for single variable functions. However, in practice we never use the inverse of a matrix for computations, so we cannot use this formula directly. Rather, we can do the following. First solve the equation Df(x0)∆x = −f(x0) Since Df(x0) is a known matrix and −f(x0) is a known vector, this equation is just a system of linear equations, which can be solved efficiently and accurately. Once we have the solution vector ∆x, we can obtain our improved estimate x1 by x1 = x0 + ∆x. For subsequent steps, we have the following process: • Solve Df(xi)∆x = −f(xi). • Let xi+1 = xi + ∆x

the polynomial x^4-2x^3 +3x^2-ax+3a-7 when divided by x+1 leaves remainder 19 .find [a]

Subtract 19: x^4-2x^3+3x^2-ax+3a-26. Because we removed the remainder, x+1 must be a factor of the revised polynomial. We can use synthetic division to divide by x+1 using -1 as the root. When we do this we end up with 4a-20. This must be equal to zero for there to be no remainder, so 4a-20=0 and a=5. -1 | 1 -2 3 -a  (3a-26) ......1 -1 3 -6  (a + 6) ......1 -3 6 (-a-6) | 0  4a-20=0, 4a=20, a=5. Polynomial was x^4-2x^3+3x^2-5x+8.

p(x)=kx^3+15x^2+9x-40 mod (x+4)=-28. find the value of 'k'.

Try different values of x to observe the behaviour of p(x). x=0: p(0)=0, because 40 mod (x+4)=40 mod 4=0. x=1: p(1)=k+24, because 40 mod 5=0. If k+24=-28, then k=-52. x=-1: p(-1)=5-k, because 40 mod 3=1. If 5-k=-28, then k=33. x=2: p(2)=8k+74, because 40 mod 6=4. If 8k+74=-28, then k=-102/8=-51/4. These give different values for k, so there is no single solution for k. However, if the question is supposed to read: p(x)=(kx^3+15x^2+9x-40) mod (x+4)=-28, the situation is quite different. Using synthetic division, we can work out the remainder: -64k+164=-28. The modulo function only requires the remainder, which we know is -28; therefore 64k=192 and k=3. Details of division: -4 | k......15..........9...........-40 ......k.....-4k 16k-60 -64k+204 ......k 15-4k 16k-51 -64k+164 (=-28)