answer:Starting with the equation [frac{x + 6}{x - 3} = 4,] perform cross-multiplication to clear the fraction: [x + 6 = 4(x - 3).] Expanding the right side: [x + 6 = 4x - 12.] Bring all the x terms to one side and constants to the other: [4x - x = 6 + 12,] [3x = 18,] Divide by 3 to solve for x: [x = 6.] Thus, the solution is [x = boxed{6}.]

answer:- **Step 1:** Rewrite s(x) and r(x) in cases: [s(x) = -|x+2| = begin{cases} x + 2 & text{if } x+2 leq 0 -(x+2) & text{if } x+2 > 0 end{cases}] [r(x) = |x+1|-3 = begin{cases} x+1-3 & text{if } x+1 geq 0 -(x+1)-3 & text{if } x+1 < 0 end{cases} = begin{cases} x-2 & text{if } x geq -1 -x-4 & text{if } x < -1 end{cases}] - **Step 2:** Compute s(r(x)). Since these are piecewise functions, check the intervals impacted by the function shifting. - **Step 3:** Evaluate s(r(x)) at each point from x=-5 to x=3 and sum the values. Calculations for each x: [x=-5, s(r(-5)) = s(-1) = -3] [x=-4, s(r(-4)) = s(-2) = -4] [x=-3, s(r(-3)) = s(-3) = -5] [x=-2, s(r(-2)) = s(-4) = -6] [x=-1, s(r(-1)) = s(-5) = -7] [x=0, s(r(0)) = s(-2) = -4] [x=1, s(r(1)) = s(-1) = -3] [x=2, s(r(2)) = s(0) = -2] [x=3, s(r(3)) = s(1) = -3] Sum: -3-4-5-6-7-4-3-2-3 = -37. Thus, the sum of the values is boxed{-37}.

answer:To find the coefficient of x^{5} in the expansion of (x^{4}+frac{1}{x^2}+2x)^{5}, we first simplify the expression inside the parentheses. We can rewrite the expression as follows: [ (x^{4}+frac{1}{x^2}+2x)^{5} = left[(x^{4}+2x+frac{1}{x^2})right]^{5} = left[({x}^{2}+frac{1}{x})^{2}+2xright]^{5} = left[({x}^{2}+frac{1}{x})^{2}right]^{5} ] This simplification leads us to consider the expansion of left({x}^{2}+frac{1}{x}right)^{10}, since left[({x}^{2}+frac{1}{x})^{2}right]^{5} = ({x}^{2}+frac{1}{x})^{10}. The general term of the binomial expansion of ({x}^{2}+frac{1}{x})^{10} is given by: [ {T}_{r+1} = {C}_{10}^{r}({x}^{2})^{10-r}left(frac{1}{x}right)^{r} = {C}_{10}^{r}{x}^{20-3r} ] To find the coefficient of x^{5}, we set the exponent of x in the general term equal to 5: [ 20 - 3r = 5 ] Solving for r: [ 20 - 5 = 3r implies 15 = 3r implies r = 5 ] Substituting r = 5 into the binomial coefficient {C}_{10}^{r} gives us the coefficient of x^{5}: [ {C}_{10}^{5} = 252 ] Therefore, the coefficient of x^{5} in the expansion of (x^{4}+frac{1}{x^2}+2x)^{5} is boxed{252}.

answer:To solve the problem, let's evaluate each option step by step: **Option A:** sqrt{(-2)^{2}} - First, calculate the square of -2: (-2)^{2} = 4 - Then, take the square root of 4: sqrt{4} = 2 So, sqrt{(-2)^{2}} = 2. Therefore, option A is incorrect because it claims sqrt{(-2)^{2}} = -2. **Option B:** -sqrt{{2}^{2}} - First, calculate the square of 2: {2}^{2} = 4 - Then, take the square root of 4: sqrt{4} = 2 - Finally, apply the negative sign: -sqrt{4} = -2 So, -sqrt{{2}^{2}} = -2. Therefore, option B is incorrect because it claims -sqrt{{2}^{2}} = 2. **Option C:** sqrt{(-2)^{2}} - This is the same calculation as in option A, where sqrt{(-2)^{2}} = 2. So, sqrt{(-2)^{2}} = 2, not pm 2. Therefore, option C is incorrect because it claims sqrt{(-2)^{2}} = pm 2. **Option D:** sqrt[3]{-8}=-sqrt[3]{8} - Calculate the cube root of -8: sqrt[3]{-8} = -2 - Calculate the cube root of 8: sqrt[3]{8} = 2 - Apply the negative sign to the cube root of 8: -sqrt[3]{8} = -2 So, sqrt[3]{-8} = -sqrt[3]{8} = -2. Therefore, option D is correct because it meets the requirements. Therefore, the correct answer is: boxed{D}.