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aev [14]
3 years ago
10

Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoo

t Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction
Mathematics
2 answers:
Licemer1 [7]3 years ago
5 0

Answer:

Using the quadratic formula to solve 5x = 6x^2 – 3, what are the values of x?

5x = 6x^2 – 3

Subtract 5x from both sides:

0 = 6x^2 – 5x – 3

a = 6, b = -5, c = -3

x = (-b ± √(b^2 - 4ac))/(2a)

x = (-(-5) ± √((-5)^2 - 4(6)(-3)))/(2(6))

x = (5 ± √(25 + 72))/12

x = (5 ± √(97))/12

Step-by-step explanation:

iris [78.8K]3 years ago
3 0

Answer:

your answer is C

Step-by-step explanation:

thank me later;)

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a. 62

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The Surface Area of this cylinder rounded to the nearest square
kipiarov [429]

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1728

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3 years ago
(3/1/2 - 9/3/4) / (-2.5)=
Alex73 [517]
<h3>Answer :</h3>

\:  \:  \:

=  \rm \large  \frac{3 \frac{1}{2}  - 9 \frac{3}{4} }{ - 2.5}

\:  \:

= \rm \large  \frac{ \frac{7}{2}  - 9 \frac{3}{4} }{ - 2.5}

\:  \:

=  \rm \large \:  \frac{ \frac{7}{2} -  \frac{39}{4}  }{ - 2.5}

\:  \:

=  \rm \large \:  \frac{ \frac{25}{4} }{ - 2.5}

\:  \:  \:

=  \rm \large \:  \frac{5}{2}

\:  \:  \:

6 0
2 years ago
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