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Brums [2.3K]
2 years ago
11

What are the exact solutions of x2 − 5x − 1 = 0 using x equals negative b plus or minus the square root of the quantity b square

d minus 4 times a times c all over 2 times a?
Mathematics
1 answer:
geniusboy [140]2 years ago
5 0

Answer:

x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}

Step-by-step explanation:

So the question here is asking you to use the quadratic formula which is expressed as: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\

A quadratic can generally be expressed as: y=ax^2+bx+c

So using the equation you gave: 0=x^2-5x-1

We can identify the following values: a=1, b=-5, c=-1

Btw the equation explicitly write "1" as the coefficient of x, but since it's not provided it's implied that it's 1.

So plugging in the known values, we get the following equation:

x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2(1)}\\\\x=\frac{5\pm\sqrt{25+4}}{2}\\\\x=\frac{5\pm\sqrt{29}}{2}\\\\x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}

The last step just consists of taking the + and - solution, and since it asks for exact solutions you leave the 29 under the radical, and you don't approximate. There is no further simplification that can be done here.

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What is the equation of the parabola?
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d=4.5t

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A realtor uses a lock box to store the keys to a house that is for sale. The access code for the lock box consists of sixsix dig
vazorg [7]

Answer:

450,000 codes

Step-by-step explanation:

Number of digits in the access code = 6

Conditions:

  • First digit cannot be 2
  • Last digit must be odd

Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.

Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.

The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.

According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.

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Hence, 450,000 different codes are possible for the lock box

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