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Lerok [7]
3 years ago
14

What are the real roots of 2x^3+9x^2+4x-15=0

Mathematics
1 answer:
klemol [59]3 years ago
8 0
Well, luckily it is apparent that (x-1) is a root because when x=1 the equation is equal to zero.  So we can divide the equation by that factor to find the other roots. 

(2x^3+9x^2+4x-15)/(x-1)
2x^2  r  11x^2+4x-15
11x    r  15x-15
15     r   0

(x-1)(2x^2+11x+15)=0

(x-1)(2x^2+6x+5x+15)=0

(x-1)(2x(x+3)+5(x+3))=0

(x-1)(2x+5)(x+3)=0

So the roots are x= -3, -2.5, 1
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Step-by-step explanation:

Given that:-

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So, we can concert 15 minutes to hours.

So,

60 minutes = 1 hour

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Okay, now we will apply first for 0.4 hours.

0.25  \: hrs= 1 \: cell

1 \: hr =  \frac{1}{0.25}

0.4 \: hrs =  \frac{0.4}{0.25}

= 1.6 \: cells

= 16  \times  {10}^{ - 1} \: cells

Now, second one:-

10 \: hrs =  \frac{10}{0.25}

= 40 \: cells

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Hope it helps :D

6 0
3 years ago
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valina [46]

Answer:

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Step-by-step explanation:

8 0
3 years ago
C. Ben and Michael also participated in the long jump. Michael jumped 334 thousandths of a meter less than Kevin. Ben jumped far
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siniylev [52]

Answer:

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8 0
3 years ago
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Ivan

Answer:

Step-by-step explanation:

Oh, this is a lot easier than you made it sound, lol.

This is the correct answer:

The table with the following:

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5 0
3 years ago
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