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marin [14]
3 years ago
14

The price of a company's share dropped by 3.50% by the end of the first year, down to $44.25. During the second year the price o

f the share dropped by $1.99. a. What was the price of the share at the beginning of the first year? Round to the nearest cent b. What was the price of the share at the end of the second year? Round to the nearest cent c. What was the percent change in the price of the share over the two years? % Express the answer with a positive sign for percent increase or negative sign for percent decrease
Mathematics
1 answer:
omeli [17]3 years ago
5 0

Answer:

a. $45.86

b. $43.87

c. Percentage change = -4.40%)

Step-by-step explanation:

a. Let the price of share at the beginning of first year be represented with X

Now, Price of company's share dropped by 3.50% at the end of first year.

So, Price of share at the end of first year = x - 3.5% of x

= x - 0.035x

= 1x - 0.035x

= 0.965x

But it is equal to $44.25

=> 0.965x = 44.25

x = 44.25 / 0.965  

x = 45.8549223

x = $45.86

b. During second year price of share decreased by $1.99. Therefore, Price of share at the end of second year = $45.86 - $1.99  = $43.87

c. Percentage change in the price of shares over two years = {(45.86 - 43.87)/45.86} *100

= (1.99/45.86)*100

= 0.04339294 * 100

= 4.40%

Now, as price of shares has dropped, the percentage change will be negative. (Δ% = -4.40%)

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Irina-Kira [14]

Label the points A,B,C

  • A = (1,2)
  • B = (4,5)
  • C = (8,9)

Let's find the distance from A to B, aka find the length of segment AB.

We use the distance formula.

A = (x_1,y_1) = (1,2) \text{ and } B = (x_2, y_2) = (4,5)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-4)^2 + (2-5)^2}\\\\d = \sqrt{(-3)^2 + (-3)^2}\\\\d = \sqrt{9 + 9}\\\\d = \sqrt{18}\\\\d = \sqrt{9*2}\\\\d = \sqrt{9}*\sqrt{2}\\\\d = 3\sqrt{2}\\\\

Segment AB is exactly 3\sqrt{2} units long.

Now let's find the distance from B to C

B = (x_1,y_1) = (4,5) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(4-8)^2 + (5-9)^2}\\\\d = \sqrt{(-4)^2 + (-4)^2}\\\\d = \sqrt{16 + 16}\\\\d = \sqrt{32}\\\\d = \sqrt{16*2}\\\\d = \sqrt{16}*\sqrt{2}\\\\d = 4\sqrt{2}\\\\

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Adding these segments gives

AB+BC = 3\sqrt{2}+4\sqrt{2} = 7\sqrt{2}

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

Now if A,B,C are collinear then AB+BC should get the length of AC.

AB+BC = AC

Let's calculate the distance from A to C

A = (x_1,y_1) = (1,2) \text{ and } C = (x_2, y_2) = (8,9)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (2-9)^2}\\\\d = \sqrt{(-7)^2 + (-7)^2}\\\\d = \sqrt{49 + 49}\\\\d = \sqrt{98}\\\\d = \sqrt{49*2}\\\\d = \sqrt{49}*\sqrt{2}\\\\d = 7\sqrt{2}\\\\

AC is exactly 7\sqrt{2} units long.

Therefore, we've shown that AB+BC = AC is a true equation.

This proves that A,B,C are collinear.

For more information, check out the segment addition postulate.

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I would substitute y = x^2

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