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Yuliya22 [10]
3 years ago
14

Which point lies on the circle represented by the equation (x+5) + (y-9) = 8

Mathematics
1 answer:
olga_2 [115]3 years ago
3 0

Answer:

(-5,9+√8)

Step-by-step explanation:

There are two ways in which you can find a point that lies in the circle. One of them is to do y the subject of the formula, and another one is to determine the center of the circumference, and with the information of the radius, you can sum this value upward or downward.

the general equation of a circle is:

(x-h)^2+(y-k)^2=r^2

with center at (h,k)

you have the following equation:

(x+5)^2+(y-9)^2=8\\\\h=-5\\\\k=9\\\\r=\sqrt{8}

Then, the center is (-5,9)

if you sum the value of the radius in one of the fourth directions (up, down, left, right), for example upward you have

(-5,9+r)=(-5,9+\sqrt{8})

Then, one point that lies in the circle is (-5,9+√8)

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a) It will take 17.71 years

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Step-by-step explanation:

a) Lets talk about the compound interest

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, Where:

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

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

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

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∵ The money deposit is $2000

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∴ P = 2000

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∴ 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)

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∴ 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

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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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