what is the answerof two powers with the same value?

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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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### 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

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### 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 soluti

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### 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

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### 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

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### 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/

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### 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

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### 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. Qu

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### 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

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### 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 coefficient

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### 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

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