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nekit [7.7K]
3 years ago
9

Simplify. answers 2x^11 4x^11 4x^2

Mathematics
1 answer:
Eddi Din [679]3 years ago
3 0

Answer:

The answer to your question is    4x¹¹\sqrt{5}

Step-by-step explanation:

Expression

                              2x\sqrt{20x^{20}}

Process

1.- Find the prime factors of 20

                   20  2

                    10  2

                     5  5

                     1

    20 = 2²5

2.- Express the square root as a fractional exponent

                             2x2²/²5¹/² x²⁰/²

3.- Simplification

                             2x¹(2)(5¹/²)(x¹⁰)

                             4x¹¹5¹/²

                             4x¹¹\sqrt{5}

4.- Conclusion

None of your answer, because we must consider \sqrt{5}.

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What's -5x – 10 = 10
Nikitich [7]
-5x-10=10   x= -4

Work:
-5x-10=10
     +10  +10
______________
-5x=20
_______
-5    -5

x=-4


3 0
3 years ago
Read 2 more answers
A plan that costs $29.99 another company $19.99 and $0.35 during nights and weekends for what numbers of night and weekend does
Masja [62]

Answer:

<em>29 minutes more</em>

Step-by-step explanation:

Let m represent minutes

changing the statement to algebra, since the second company charges a different rate at night and weekend  we have the equation below;

$19.99 + $0.35m > $29.99

Subtract 19.99 from both sides to isolate m and we have;

$19.99 -$19.99 + $0.35 > $29.99 - $19.99

= $0,35m > $10.00

Divide both side by 0.35 to obtain the value of m;

\frac{0.35m}{0.35} > \frac{10}{0.35}

= m > 28.57

<em>m ⩾ 29 minutes</em>

<em>The second company's will be twenty nine minutes or more costlier than the first company</em>

5 0
3 years ago
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A box of fruit has three times as many apples as oranges
luda_lava [24]
What is the question? i could help you if i knew the question
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3 years ago
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erastovalidia [21]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
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The sum of the squares of two consecutive positive integers is 41. Find the two
Komok [63]

We have to present the number 41 as the sum of two squares of consecutive positive integers.

1² = 1

2² = 4

3² = 9

4² = 16

5² = 25

16 + 25 = 41

<h3>Answer: 4 and 5</h3>

Other method:

n, n + 1 - two consecutive positive integers

The equation:

n² + (n + 1)² = 41     <em>use (a + b)² = a² + 2ab + b²</em>

n² + n² + 2(n)(1) + 1² = 41

2n² + 2n + 1 = 41     <em>subtract 41 from both sides</em>

2n² + 2n - 40 = 0     <em>divide both sides by 2</em>

n² + n - 20 = 0

n² + 5n - 4n - 20= 0

n(n + 5) - 4(n + 5) = 0

(n + 5)(n - 4) = 0 ↔ n + 5 = 0 ∨ n - 4 =0

n = -5 < 0 ∨ n = 4 >0

n = 4

n + 1 = 4 + 1 = 5

<h3>Answer: 4 and 5.</h3>
7 0
3 years ago
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