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lbvjy [14]
3 years ago
13

What is the largest number c such that 2x^2 + 5x + c = 0 has at least one real solution? Express your answer as a common fractio

n
Mathematics
2 answers:
Vlad [161]3 years ago
8 0

Answer:

2x^2  - 5x  +  k  =   0

Step-by-step explanation:

pashok25 [27]3 years ago
4 0
I believe it should be 25/8
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Write a real world problem for: f + 5 = 9
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9-5=4 Therefore f=4
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the Marked price of an article is rupees 1600 if it is sold for rupees 1420 what is the discount rate​
Naya [18.7K]

Answer:

1600-1420= 180

Step-by-step explanation:

Given that

The Marked price of an article is rupees 1600 if it is sold for rupees 1420 then what is the discount rate

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1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

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Add 1/2 to both sides:

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Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

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It costs $859.32 to have a school dance (tickets $8)
djverab [1.8K]

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