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Dvinal [7]
3 years ago
6

3x^2 + 64 = 2x^2/2 find the roots of the quadratic equation

Mathematics
1 answer:
Reil [10]3 years ago
8 0

Answer:

4√2i , -4√2i.

Step-by-step explanation:

3x^2 + 64 = 2x^2/2

3x^2 + 64 = x^2

2x^2 + 64 = 0

2x^2 = -64

x^2 = -32

x = +/- √-32

x = +/- √16* -2

x = +/- 4√2 i.

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\frac{AC}{BC} =  \frac{DF}{EF}
Substitute for the values of AC, BC, DF and EF:
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x = 5.5 \: units
To solve for y, do the same thing. Pick two sides on triangle ABC and their corresponding sides in triangle DEF.
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3241004551 [841]

Answer:

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Step-by-step explanation:

1)

Final amount (A) = 2400 ; rate (r) = 6% = 0.06, time, t = 1.5 years

Sum = principal = p

Using the relation :

A = p(1 + rt)

2400 = p(1 + 0.06(1.5))

2400 = p(1 + 0.09)

2400 = p(1.09)

p = 2400 / 1.09

p = 2201.8348

2.)

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15600 = 12000(1 + 0.1t)

15600 = 12000 + 1200t

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t = 3600 / 1200

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3.)

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x = p(1 + x/100 * 1/x)

x = p(1 + x /100x)

x = p(1 + 1 / 100)

x = p(1 + 0.01)

x = p(1.01)

x / 1.01 = p

x / (1 + 0.01)

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3 years ago
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Answer:

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Step-by-step explanation:

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