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Sedbober [7]
3 years ago
5

A 20,000 deposit earns 4.8 interest for 4 years . What is the final account balance if the interest is compounded semiannually ?

Quarterly? Monthly ?
Mathematics
1 answer:
Bad White [126]3 years ago
4 0

Answer:

Semi-annually: A = $24 178.51

Quarterly:         A = $24 205.73

Monthly:           A = $24 224.13

Step-by-step explanation:

The formula for compound interest is

A = P(1 + r)ⁿ

A. Compounded semi-annually

Data:

P = $20 000

APR  = 4.8 %

t = 4 yr

Calculations:

n = 4 × 2 = 8

r = 0.048/2  = 0.024

A = 20 000(1+ 0.024)⁸

  = 20 000 × 1.024⁸

  = 20 000 × 1.208 926

  = $24 178.51

B. Compounded Quarterly

n = 4 × 4 = 16

r = 0.048/4  = 0.012

A = 20 000(1+ 0.012)¹⁶

  = 20 000 × 1.012¹⁶

  = 20 000 × 1.210 286

  = $24 205.73

C. Compounded monthly

n = 4 × 12 = 48

r = 0.048/12  = 0.004

A = 20 000(1+ 0.004)⁴⁸

  = 20 000 × 1.004⁴⁸

  = 20 000 × 1.211 207

  = $24 224.13

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Answer:

The y-value of the vertex is -12

Step-by-step explanation:

we know that

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where

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In this problem we have

f(x)=4x^{2}+8x-8 -----> this a vertical parabola open upward

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

f(x)+8=4x^{2}+8x

Factor the leading coefficient

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Complete the square. Remember to balance the equation by adding the same constants to each side

f(x)+8+4=4(x^{2}+2x+1)

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Rewrite as perfect squares

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The vertex is the point (-1,-12)

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3 years ago
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zhuklara [117]

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Step-by-step explanation:

3 0
3 years ago
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Hi guys, can anyone help me with this triple integral? Many thanks:)
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Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

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Let's work on the inner one,

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There's no z in the integrand, so we treat it as a constant.

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So the middle integral is

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3 years ago
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The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If
stira [4]
The is the concept of algebra, the weight of the spring can be modeled by the equation;
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3 years ago
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Answer:

The first one is your answer.

Step-by-step explanation:

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Now you look at the x^5. What you do is you make it until you have 5 x's like this x, x, x, x, x. You then put them in pairs of two.

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