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svlad2 [7]
3 years ago
14

What is the future value of a R 75 000 deposit made every 6 months for 3 years using an annual rate of 10% compounded semi-annua

lly?​
Mathematics
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

The future value is: RS 132867.075

Step-by-step explanation:

Given

PV = 75000 --- present value

r = 10\% --- annual rate

n = 6 --- number of times compounded (3 years/ 6 months)

Required

The future value

This is calculated as:

FV =PV * (1 + r)^n

FV = 75000 * (1 + 10\%)^6

FV = 75000 * (1 + 0.10)^6

FV = 75000 * (1.10)^6

FV = 132867.075

<em>Hence, the future value is: RS 132867.075</em>

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Find the amplitude and the equation of the midline of the periodic function.
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Answer:

Option B. Amplitude =3 midline is y =2.

Step-by-step explanation:

In the graph attached we have to find the amplitude and midline of the periodic function.

Amplitude of the periodic function = (Distance between two extreme points on y asxis)/2

=  (5-(-1))/2 = (5+1)/2 =6/2 =3.

Since amplitude of this function is 3 and by definition amplitude of any periodic function is the distance between the midline and the extreme point of wave on one side.

Therefore midline of the wave function is y=2 from which measurement of the amplitude is 3.

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Step-by-step explanation:

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dexar [7]

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

  • width = w
  • length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

----------------------

Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150  being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

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