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# what is the answerof two powers with the same value?

two powers that have the same value or answer it can be a different number with a different exponet be ut the same value

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## Suggested Questions And Answer :

### what is the answerof two powers with the same value?

1 raesed tu NE power is 1 0 raesed tu NE power is 0, sept that 0^0 not defined (NE number) raesed tu 0 power=1, sept 0^0

### In algebra, why are all odd powers reversible?

Strictly speaking "NO" power operation (raising to an exponent) is reversible, because the inverse operation is ALWAYS multi-valued, unless we restrict the domain. In the case of even powers, for example: (+3) ^2 = 9 (- 3) ^2 = 9 Therefore the reverse operation thus has two possible solutions: Sqrt(9) = +3 Sqrt(9) =  -3 Also, it is "NOT" true that odd powers are reversible, if we consider complex numbers: (+3                        ) ^3  = 27 (-1.5 + j 2.59808) ^3  = 27 (-1.5 -  j 2.59808) ^3  = 27 The cube root of three (above) therefore has a total of three solutions, one "real"  and two complex conjugates. In the general case, the inverse operation of: y = (x)^n        --->   x = nth root of (y) has "n" solution. Odd roots(y) have a single real solution and (n-1)/2 pairs of complex conjugate solutions. Even roots of (y) have two real solutions (+/-) and (n-2)/2 pairs of complex conjugate solutions, assuming that "y" is positive. In the general case, the "nth root" of any real or complex power can be plotted as a set of equally spaced vectors in the complex plane, each with a magnitude of:    (|y|  / n) separated by angles of:   (2 pi / n)   or    (360 degrees /n)   In summary:     Even powers are reversible if we restrict our domain to positive numbers.     Even powers are NOT reversible in the INTEGER or larger domains,     because the roots are multi valued.     Odd powers are reversible in the REAL domain,     but NOT in the COMPLEX or larger domains. Hope this helps, MJS

### Find x to the 3 power - 2y to the 2 power -3z to the 3 power + z to the 4 power if z = 3, y = 5, and z = -3

Find x to the 3 power - 2y to the 2 power -3z to the 3 power + z to the 4 power if z = 3, y = 5, and z = -3 I'm assuming there's an error here, where it says :- z = 3, y = 5, and z = -3 Since you have two values for z and none for x then I'm going to take the 1st value for z as the actual value for x, So your question should read as, Find x to the 3 power - 2y to the 2 power -3z to the 3 power + z to the 4 power if x = 3, y = 5, and z = -3   Rewriting x to the 3 power - 2y to the 2 power -3z to the 3 power + z to the 4 power as x^3 - 2y^2 - 3z^3+ z^4, Substituting for x = 3, y = 5, and z = -3 into x^3 - 2y^2 - 3z^3+ z^4, we get 3^3 - 2(5)^2 - 3(-3)^3 + (-3)^4 27 - 2(25) - 3(-27) + (9)^2 27 - 50 + 81 + 81 27 + 162 - 50 27 + 112 139 Answer: 139

### m and n positive integers, m‹n.expansion of (1+2x)power n, coefficient of x power m is 3/2 times coefficient of x power m-1,show that m and n satisfy 4n-7m+4=0.find smallest values of m and n

(1+2x)^n=1+2nx+n(n-1)/2*4x^2+n(n-1)(n-2)/6*8x^3+...+n(n-1)(...)(n-m+1)/m!(2x)^m (up to mth term). If we take two consecutive terms m-1 and m, a[m-1]x^(m-1) and a[m]x^m, we can relate the terms: mth term: n(n-1)(...)(n-m+1)/m!(2x)^m (m-1)th term: n(n-1)(...)(n-m)/(m-1)!(2x)^(m-1) mth term/(m-1)th term=(n-m+1)/m * 2x=3/2. (2x)^m=2^m*x^m and (2x)^(m-1)=2^(m-1)*x^(m-1). So there is a factor of 2 between two consecutive powers of x. Therefore 2(n-m+1)/m=3/2; 4n-4m+4=3m; 4n-7m+4=0 QED. Therefore m=(4n+4)/7 so 4n+4=4(n+1) has to be a multiple of 7, i.e., n+1 must be a multiple of 7, so n=6 is the lowest value, and m=4. CHECK Coefficients when n=6 are 1 6 15 20 15 6 1 (coefficients of (2x)), and m starts from 0 because (2x)^0=1, the first term. When we include the factor of 2 we get: 1 12 60 160 240 192 64. When m=4 we have 160 and 240 as consecutive terms and 240/160=3/2.

### Binomial Expansion- Core Maths

Question: The first two terms in the expansion of (1+ax/2)^10 + (1+bx)^10 in ascending powers of x., are 2 and 90x^2  Given that a is less than b find the values of the constants a and b . (11 marks) The Binomial expansions of the two expressions are as follows, (1+ax/2)^10 = 1 + 10.(ax/2) + 10.9/2.(ax/2)^2 + ... (1+bx)^10 = 1 + 10.(bx) + 10.9/2.(bx)^2 + ... Sum = 2 + 10.(ax/2 + bx) + 10.9/2.(a^2x^2/4 + b^2x^2) + ... Sum = 2 + 10.(a/2 + b)x + 45(a^2/4 + b^2)x^2 + ... Given information is that Sum = 2 + 90x^2 + ..., hence 10(a/2 + b) = 0   --> a = -2b 45(a^2/4 + b^2) = 90 i.e. a^2/4 + b^2 = 2 4b^2/4 + b^2 = 2   -- using a = -2b 2b^2 = 2 b^2 = 1 b = +/- 1 If b = -1, then a = 2 if b = +1, then a = -2 Since we are given that a < b, then a = -2 and b = 1. Answer: a = -2, b = 1

### complex, rational and real roots

The quintic function should have 5 roots.  The changes of sign (through Descartes) tell us the maximum number of positive roots. Since there are two changes of sign there is a maximum of 2 positive roots. To find the number of negative roots we negate the terms with odd powers and check for sign changes: that means that there is at most one negative root, because there's only one sign change between the first and second terms. Complex roots always come in matching or complementary pairs, so that means in this case 2 or 4.  Put x=-1: function is positive; for x=-2 function is negative, so there is a real root between -1 and -2, because the x axis must be intercepted between -1 and -2. This fulfils the maximum for negative roots. That leaves 4 more roots. They could be all complex; there could be two complex and two positive roots. Therefore there are at least two complex roots. To go deeper we can look at calculus and a graph of the function: The gradient of the function is 10x^4-5. When this is zero there is a turning point: x^4=1/2. If we differentiate again we get 40x^3. When x is negative this value is negative so the turning point is a maximum when we take the negative fourth root of 1/2; at the positive fourth root of 1/2 the turning point is a minimum, and these are irrational numbers. The value of the function at these turning points is positive. The fourth root of 1/2 is about 0.84. and once the graph has crossed the x axis between -1 and -2, it stays positive, so all other roots must be complex. The function is 12 when x=0, so we can now see the behaviour: from negative values of x, the function intersects the x axis between -1 and -2 (real root); at x=4th root of 1/2 (-0.84) it reaches a local maximum, intercepts the vertical axis at 12 until it reaches 4th root of 1/2 (0.84) and a local minimum, after which it ascends rapidly at a steep gradient.  [Incidentally, one way to find the real root is to rearrange the equation: x^5=2.5x-6=-6(1-5x/12); x=-(6(1-5x/12))^(1/5)=-6^(1/5)(1-5x/12)^(1/5) We can now use an iterative process to find x. We start with x=0, so x0, the first approximation of the solution, is the fifth root of 6 negated=-1.430969 approx. To find the next solution x1, we put x=x0 and repeat the process to get -1.571266. We keep repeating the process until we get the accuracy we need, or the calculator reaches a fixed value. After just a few iterations, my calculator gave me -1.583590704.]

### how do you write an equation for the linear function f with the given values?

Standard linear function is f(x)=ax+b where a and b are constants. Substitute each point into the function: 21=-2a+b, -35=5a+b. We need go no further because we have two linear equations and two unknowns. f(-2)-f(5): 56=-7a so a=-8. Now we can work out b: b=21+2a=21-16=5 so f(x)=5-8x. Quick check shows that the function is correct: f(-2)=5+16=21; f(5)=5-40=-35. But we need to check all the other points: f(-6)=5+48=53; f(3)=5-24=-19; uh-oh, something wrong! The values don't fit. So either f(x) is not linear or it's piecewise. Closer inspection is needed. First plot the points. It's clear to see that the points are not colinear. Join the points with straight lines. We're not told that f is continuous. Let's assume it is. If it's piecewise, we need 5 different equations to define the function between the 6 points. In order these are (-9,-4), (-6,-2), (-2,21), (3,-5), (5,-35), (12,14). Call these points A, B, C, D, E, F. We can work out linear equations between A and B, B and C, C and D, etc. These equations will provide continuity for f in the domain -9 Read More: ...

### Prove that x^12 - y^12 is divisible by 1365.

The 12th power of all numbers (integers) that are not divisible by 5 end in the digits 1 or 6. So when the 12th powers of such numbers are subtracted from one another the result will always end in 0 (from 2 1's or 2 6's) or 5 (from 1-6 or 6-1). Therefore this result is divisible by 5. If one of the numbers is a multiple of 5 its 12th power will end in 5, so if the other number is not divisible by 5 its 12th power will end in 1 or 6 so the difference cannot be divisible by 5. Since 5 is a factor of 1365, it cannot be generally stated that x^12-y^12 is divisible by 1365 since it has just been shown that if x or y (not both) is divisible by 5, the difference is not divisible by 5 and therefore cannot be divisible by 1365. So the statement is not true for all values of x and y. 1365 factorises: 3*5*7*13.  Numbers not divisible by 7 have differences of 6th powers that are divisible by 7, so differences of 12th powers will also be so divisible. The difference between the 12th powers of any two integers neither of which are divisible by 5 or 7 will be divisible by 35. There are similar rules of divisibility for 3 and 13, but for the statement to be true constraints have to be applied to x and y, and these will exclude divisibility by the factors of 1365. So the statement is generally false.

### How to find the factors of a 7th degree polynomial

If there are only positive coefficients, the zeroes (from which factors can be deduced) will all be negative. For mixed positive and negative coefficients, usually there are some simple zeroes like 1 and -1 that can be substituted for the variable that give a zero result. The size of the coefficients are also likely to be small numbers. Look out for missing powers; that may be an indication that the polynomial has a factor involving x^2 rather than x, for example. When you find a factor divide the polynomial by it using algebraic or synthetic division. This will reduce the degree. The method I find to be most effective is to use my calculator to plug in trial values for the variable between, say, -5 and +5 in steps of 1. When the result changes sign between two consecutive integers, I know there's a zero between; and occasionally an integer will itself produce zero, so I know that integer is a zero. Once a zero is located between two integers I can then home in on values in between. In a 7-degree poly you may have up to 7 factors so you do have to keep looking. Once you have found 5 factors using zeroes (if a is a zero then x-a is a factor), you will be left with a quadratic which may or may not factorise further, but you can apply the formula in any case. I hope this helps.

### Estimate: ((36,421 x 10^5)(493,025))/40,216 x 10^7

Estimate means work out the answer roughly to give a good idea of what sort of value the true answer will be. This is to provide some check on the validity of the answer when properly calculated. I would note that the denominator 40216 has a similar size to 36421, so that leaves two powers of 10 which gives us 10^-2 or a hundredth, which can be applied to 493025, a number close to 500000. Take two zeroes off the end of this to give 5000 (the result of dividing by 100). So the answer is roughly 5000. When I calculate the answer I can expect a value around 5000. Let's see: the answer is 4465.004855 which is of the sane order as 5000. I could simply have removed 25 from 493025 giving me 4930, an even closer estimate.