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Published on November 13, 2024
To find (a+b)^2, given a^2 + b^2 = 29 and ab = 10, substitute the given values into the equation (a+b)^2 = a^2 + 2ab + b^2, resulting in (a+b)^2 = 49.
The only real solution to the cubic equation a^3 + a^2 - 36 = 0 is a = 3.
The solution to the equation X^3 - X^0 = 26 is X = 3.
The equation (x-1)^(|x|-4) = 1 has solutions x = -4, 0, 2, and 4.
The equation x^2 - 2x = 0 has two solutions, x = 0 and x = 2, not just x = 2.
The solution for b in the equation 2^(b^2+2) = 64 is b = 2.
The multiplication of the two given matrices results in a 2x3 matrix with entries calculated by multiplying corresponding rows and columns of the matrices and summing the products.
The equation x^x + 6/x = 7 has two solutions, x = 1 and x = 2, as demonstrated by direct substitution.
The solution to the equation 3^(x+2) = 2^(x+3) is found by taking the natural logarithm of both sides, applying logarithm properties, isolating x, and simplifying to an expression involving natural logarithms of 2 and 3. The approximate value of x is then calculated.
The expression $((i+1)(i-1))^2$ simplifies to 4.
The equation 3^(x^2)/9^x = 81 is solved by rewriting the equation with the same base, simplifying the denominator, applying the quotient rule for exponents, equating the exponents, rearranging into a quadratic equation, and finally using the quadratic formula to find the two solutions, x = 1 + √5 and x = 1 - √5.