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poizon [28]
3 years ago
14

What is the line segments between (4,-6) and (-6,-16)?

Mathematics
1 answer:
sergejj [24]3 years ago
6 0
The line segments between (4,-6) and (-6,-16) is 1.

slope = y2 -y1 /x2-x1 = -16+6/-6-4=-10/-10=1
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Please help this for my pre calculus finals
dsp73

Answer:

a) It will take 17.71 years

b) It will take 17.58 years

c) I will earn $6.60 more in compound continuously

Step-by-step explanation:

a) Lets talk about the compound interest

- The formula for compound interest is A = P (1 + r/n)^(nt)

, Where:

- A = the future value of the investment, including interest

- P = the principal investment amount (the initial deposit)

- r = the annual interest rate (decimal)

- n = the number of times that interest is compounded per unit t

- t = the time the money is invested

* Lets solve the problem

∵ The money deposit is $2000

∵ The rate is 6.25%

∵ The interest is compound quarterly

∵ The future value is $6000

∴ P = 2000

∴ A = 6000

∴ r = 6.25/100 = 0.0625

∴ n = 4

∴ t = ?

∵ A = P (1 + r/n)^(nt)

∴ 6000 = 2000 (1 + 0.0625/4)^4t ⇒ divide both sides by 2000

∴ 3 = (1.015625)^4t ⇒ insert ㏑ for both sides

∴ ㏑(3) = ㏑(1.015625)^4t

∵ ㏑(a)^b = b ㏑(a)

∴ ㏑(3) = 4t ㏑(1.015625) ⇒ divide both sides by ㏑(1.015625)

∴ 4t = ㏑(3)/㏑(1.015625) ⇒ divide both sides by 4

∴ t = [㏑(3)/㏑(1.015625)] ÷ 4 = 17.71

* It will take 17.71 years

b) Lets talk about the compound continuous interest  

- Compound continuous interest can be calculated using the formula:

  A = P e^rt  

- A = the future value of the investment, including interest

- P = the principal investment amount (the initial amount)

- r = the interest rate  

- t = the time the money is invested

* Lets solve the problem

∵ The money deposit is $2000

∵ The rate is 6.25%

∵ The interest is compound continuously

∵ The future value is $6000

∴ P = 2000

∴ A = 6000

∴ r = 6.25/100 = 0.0625

∴ t = ?

∵ A = P e^rt  

∴ 6000 = 2000 e^(0.0625 t) ⇒ divide both sides by 2000

∴ 3 = e^(0.0625 t) ⇒ insert ㏑ to both sides

∴ ㏑(3) = ㏑[e^0.0625 t]

∵ ㏑(e^a) = a ㏑(e) ⇒ ㏑(e) = 1 , then ㏑(e^a) = a

∴ ㏑(3) = 0.0625 t ⇒ divide both sides by 0.0625

∴ t = ㏑(3)/0.0625 = 17.5778

* It will take 17.58 years

c) If t = 5 years

# The compound quarterly:

∵ A = P (1 + r/n)^(nt)

∴ A = 2000 (1 + 0.0625/4)^(4×5)

∴ A = 2000 (1.015625)^20 = $2727.08

# Compound continuously

∵ A = P e^(rt)

∴ A = 2000 e^(0.0625×5) = $2733.68

∴ I will earn = 2733.68 - 2727.08 = $6.60

* I will earn $6.60 more in compound continuously

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Answer:

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