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The only appropriate response I can think of is how to interpret the exponent Any parentheses raised to the nth power means you have to rewrite that expression n times and then multiply them all In this case the exponent is just 2, so we rewriFind the value of the constant k that makes fxy(x,y) a valid joint PDF 0 Proving that $\Gamma(\frac{1}{2})=\sqrt(\pi)$ using the expected value of standard5 Convierta las siguientes fracciones a números decimales a) – ½ b) 7/3 c) 12/5 d) 100/101 ) Recibe ahora mismo las respuestas que necesitas!

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Simplify (abc)(abc) Expand by multiplying each term in the first expression by each term in the second expression Simplify terms Tap for more steps Combine the opposite terms in Tap for more steps Reorder the factors in the terms and Subtract fromSolve for c a=1/2*((bc)h) Rewrite the equation as Multiply both sides of the equation by Simplify Tap for more steps Cancel the common factor of Tap for more steps Cancel the common factor Rewrite the expression Multiply by Apply the distributive propertyIf a^x=b b^y=c c^z=a Prove xyz=1 Please give the solution a x =b => x=log a b b y =c=> y=log b c c z =a=> z=log c a xyz=(log a blog b c)log c a =log

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An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i 2 = −1 The square of an imaginary number bi is −b 2For example, 5i is an imaginary number, and its square is −25By definition, zero is considered to be both real and imaginary Originally coined in the 17th century by René Descartes as aNov 02, 07 · What an innocently looking equation If we allow negative integers, a=4, b=1, c=11 is a solution Do some tricks with divisibility?Conversely, assume that f(f−1) = C for all C ⊆ Y but that f is not ontoThen there exists y∈ Y such that for all x∈ X, we have y6= f(x) Let C= {y} Then f−1 = ∅ and f(f−1) = ∅ 6= C since y∈ C ContradictionHence, f is onto Thus, f is onto if and only if f(f−1) = C for all C⊆ Y 8(a) Claim f(A∪B) = f(A) ∪f(B)

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Considering y = 1 as constant The expression becomes a quadratic in x If it forms a perfect square, then product of roots should be 1, since x = 1 is a solution Hence, a (b − c) c (a − b) = 1 a c − b c = a b − a c 2 a c = a b b c b 2 = c 1 a 1 Hence, a 1 , b 1 , c 1 are in AP Hence, a, b, c are in HPIt won't help There are solutions, but the smallest solution has 80 digits What happens if» ¶ È( ß f N É æ à ý S631) 2 f g J G ¶ » ¶

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