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question:Cobra Banking Group issued preferred shares 10 years ago with a stated value of 95 and a dividend yield of 7%. What is the current price of each preferred share if the required return today is 15%?
answer:To calculate the current price of each preferred share, we can use the formula for the present value of a perpetuity, which is: PV = C / r where: PV is the present value C is the annual cash flow (in this case, the dividend payment) r is the discount rate or required return In this case, the annual dividend payment is: Dividend = Stated value × Dividend yield = 95 × 7% = 6.65 And the required return is 15%, or 0.15. Substituting these values into the formula, we get: PV = 6.65 / 0.15 = 44.33 Therefore, the current price of each preferred share is 44.33.
question:How many plants are in Jillian's garden if there are 3 rows of carrots, 2 rows of string beans, and 1 row of peas, with each row containing 8 plants?
answer:To calculate the total number of plants, we need to multiply the number of plants in each row by the number of rows for each type of vegetable and then add the results together. 3 rows of carrots × 8 plants per row = 24 plants 2 rows of string beans × 8 plants per row = 16 plants 1 row of peas × 8 plants per row = 8 plants Total number of plants = 24 + 16 + 8 = 48 plants Therefore, there are 48 plants in Jillian's garden.
question:Find the coefficients of x^0, x^1, x^2, and x^3 in the Maclaurin series of the function f(x)=(5x^3)e^{5x}.
answer:Series for exponential function is given by, {eq}begin{align} displaystyle e^x &=sum_{n=0}^{infty} frac{x^n}{n!} displaystyle e^{5x} &=sum_{n=0}^{infty} frac{(5x)^n}{n!} &&....text{Substituting 5x in place of x} displaystyle (5x^3)e^{5x} &=(5x^3)sum_{n=0}^{infty} frac{(5x)^n}{n!} &&....text{Multiplying by }5x^3 displaystyle (5x^3)e^{5x} &=sum_{n=0}^{infty} frac{5^{n+1}x^{n+3}}{n!} displaystyle (5x^3)e^{5x} &=frac{5^{0+1}x^{0+3}}{0!} + frac{5^{1+1}x^{1+3}}{1!} + frac{5^{2+1}x^{2+3}}{2!} + frac{5^{3+1}x^{3+3}}{3!} + ... &&....text{Expanding the summation} displaystyle (5x^3)e^{5x} &=5x^3 + 25x^4 + frac{125}{2}x^5 + frac{625}{6}x^6 ... displaystyle sum_{n=0}^{infty} c_nx^n &=5x^3 + 25x^4 + frac{125}{2}x^5 + frac{625}{6}x^6 +... displaystyle color{blue}{c_0=0, c_1} &color{blue}{=0, c_2=0, c_3=5, c_4=25} &&....text{Comparing corresponding coefficients} end{align} {/eq}
question:Factor the following quadratic: -6 x^2 + 72 x - frac{357}{2}
answer:To factor the quadratic, we can first find two numbers that add up to 72 (the coefficient of the x-term) and multiply to -frac{357}{2} (the constant term). These numbers are 81 and -frac{9}{2}. We can then rewrite the quadratic as: -6 x^2 + 72 x - frac{357}{2} = -6(x^2 - 12x + frac{9}{2}) Next, we can complete the square inside the parentheses by adding and subtracting left(frac{12}{2}right)^2 = 36 to the expression: -6(x^2 - 12x + 36 - 36 + frac{9}{2}) Simplifying this expression, we get: -6[(x - 6)^2 - frac{63}{2}] Finally, we can factor out a -6 from the expression to get: -6(x - 6)^2 + frac{63}{2} -6(x - 6)^2 + frac{63}{2} = -6(x - 6)^2 + 31.5 -6(x - 6)^2 + 31.5 = -6(x - 6)^2 + 31.5(1) -6(x - 6)^2 + 31.5 = -6(x - 6)^2 + (3sqrt{10.5})^2 -6(x - 6)^2 + (3sqrt{10.5})^2 = (3sqrt{10.5} - (x - 6))^2 Therefore, the factors of the quadratic are: (3sqrt{10.5} - (x - 6)) and (3sqrt{10.5} + (x - 6)) Simplifying these expressions, we get: (3sqrt{10.5} - x + 6) and (3sqrt{10.5} + x - 6) Multiplying these two expressions together, we get: (3sqrt{10.5} - x + 6)(3sqrt{10.5} + x - 6) Expanding this product, we get: 9(10.5) - 6(3sqrt{10.5})x + 6(3sqrt{10.5})x - x^2 + 36 Simplifying this expression, we get: -x^2 + 72x - frac{357}{2} Therefore, the factors of the quadratic are: -6(x - frac{17}{2})(x - frac{7}{2}) The answer is -6(x - frac{17}{2})(x - frac{7}{2})