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dezoksy [38]
3 years ago
8

Solve quadratic equation x²-8x-1=0 by formula method​

Mathematics
2 answers:
Alex_Xolod [135]3 years ago
6 0

Answer:

x²-8x-1=0

comparing above equation with ax²+bx+c=0

a=1

b=-8

c=-1

x=

x =  \frac{ - b +  -  \sqrt{ {b}^{2} - 4ac } }{2a}

=(--8+-√(64-4×1×-1)/2×1

=8+-√(64+4)/2

taking positive

x=(8+√68)/2=2(4+√17)/2=4+√17

taking negative

x=(8-√68)/2=2(4-√17)/2=4-√17

Ad libitum [116K]3 years ago
3 0

maybe be it's right.....

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8 0
3 years ago
Alexander can earn money for the cans he recycles. Which of the following statements describes the variables in this situation c
Murrr4er [49]
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8 0
3 years ago
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Please help me with this!!!
Dafna1 [17]

Answer:

option C is correct answer ..

Step-by-step explanation:

angle A + angle B + angle C = 180° ( by angle sum property of triangle )

3x + 4x-19+ 3x -1 = 180

10x + -20 = 180

10x = 180 +20 = 200

x = 200/10 = 20 °

angle A = 3× 20 = 60 °

angle B = 4× 20 - 1 9 = 61 °

angle C = 3× 20 -1 = 59 °

angle B is greatest so side opposite to it will be greatest in length ....so length of AC is greatest ....

so option C is the correct answer of this question ...

plz mark my answer as brainlist plzzzz vote me also as I have done this question by taking a lot of time Hope it will be helpful for you ....

5 0
3 years ago
Which expression represents the product of negative three-fourths x and –8x? negative 35 over 4 x 6x negative 35 over 4 x square
valentinak56 [21]

The expression that represents the product of negative three-fourths x and –8x is 6x^2

<h3>How to determine the equivalent expression?</h3>

The statement is given as:

the product of negative three-fourths x and –8x

Rewrite properly as:

-3x/4 * -8x

Divide 4 and 8 by 4

-3x * -2x

Multiply -3 and -2

6x * x

Multiply x and x

6x^2

Hence, the expression that represents the product of negative three-fourths x and –8x is 6x^2

Read more about expressions at:

brainly.com/question/723406

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2 years ago
9. The equation y = x2 - 1 is
SashulF [63]
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y = x²-1 is nonlinear and nonproportional.
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