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Variables and Squares, See the working below for the problem. Solve: 3√(6𝑥 + 9) + 8 = 5 Solution Answers: x= -6. Work: 3√6𝑥 + 9 + 8 = 5 3√6𝑥 + 9 = 5–8 3√6𝑥 + 9 = –3. We cube both sides 6 𝑥 +9 = -27 6 𝑥 = –36 𝑥 = –6 Variables and […]
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Squares and square roots: Equations Find the real value(s) of x that verify: √𝑥 + √7𝑥 + 2 = 0 solution Answer: NO REAL VALUES. Work: Let √𝑥 be a and √ (7x +2) be b If a>=0, b>=0 then√𝑥 (a) + √ (7x +2)(b)=0, Then a=b=0, for this equation to be
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Basic math: Mathematics, Expressions, Indices and Logarithms Evaluate the expression: 𝑙𝑜𝑔82 solution Answers: x =1/3 Work: rewrite the log to express the unknown value of x Log82 =x Remove the log, 8x=2 the we solve this by reducing 8 to base 2. Therefore, 8x=2 becomes 2(3)x =21 . the bases are the same
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Problem 3: Solve the equation: (1 − 𝑥)^3/2 = 8 solution Answer: x= -3 Work: square both sides of the equation. (1-x)3 =64. Then we get the cube root on both sides. (1-x) = 4, we then solve to get the value of x. 1–x = 4. collect the
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Question: Determine the inverse of the function: ℎ(𝑥) = 2𝑥−3/(1−𝑥) Solution and steps followed Answers: f-1 (x)= x+3 2+x Work: we replace h(x) with y, y= 2x-3 1-x We then swap the y for x, x = 2y–3 1-x Solve for y, x(1–y) = 2y–3 x-xy= 2y–3 2y+xy = x +3.
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the solution below is from differential problem
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how to calculate vectors and Eigen values Solution
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Solution :Matrices and law of matrices and Squares Question: 1 Since we are told that A is indepented, then we have A2 = A (I – A )2 we expand the problem and then solve (I – A)2 = I2 + A2 – 2AI2 (I– A)2 = I + A –2A
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