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quester [9]
3 years ago
8

Joe Hernandez has inherited $25,000 and wishes to purchase an annuity that will provide him with a steady income over the next 1

2 years. He has heard that the local bank is currently paying 6% p.A. (compounding annually). If he were to deposit his fund, what is the equal amount would he be able to withdraw at the end of each year for 12 years?
Mathematics
2 answers:
mote1985 [20]3 years ago
8 0

Answer:

p=2982

Step-by-step explanation:

Rate of interest = 6 % p.a

No of year = 12

P.V = $ 25000

Installments (p) = ?

We know the formula of P.V annuity

P.V = P[ 1-(1+r)-n/r]

25000 = P[1-(1+0.06)-12/0.06]

P = 2982 will withdraw annuity

elena-14-01-66 [18.8K]3 years ago
7 0
2 always the same number
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Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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I WILL GIVE BRAINLIEST IF U CAN HELP ME WITH ANY OF MY QUESTIONS!!!
faltersainse [42]

Answer:

12.5

Step-by-step explanation:

square root of 10^2 - 8^2 = 6 gives us missing side

sin^-1(8/10) = 53.13 gives us half of the angle

90-53.13 = 36.869 gives us the angle of the triangle on the right

tan(36.869) x 6 = 4.5

4.5 + 8 = 12.5

Hope that helps,

Good Luck! x

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