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juin [17]
3 years ago
11

Simplify square root 108x^5y^6

Mathematics
1 answer:
umka21 [38]3 years ago
8 0
√<span>108x^5y^6
First, break up </span><span>√108 :
</span>√108 = √4 x √27 = 2 x √3 x √9 = 2 x √3 x 3 = 6<span>√3
</span><span>Since the x^5 is under a root of 2, that means we can take out an x^2 and leave one x under the radical :
</span>√x^5 = x^2 (<span>√x)
</span><span>Since the y^6 is raised to an even power and the root is even (2), that means we can take out all of the y's without leaving any under the radical :
</span><span>√y^6 = y^3
</span><span>Now, combine all of our simplified forms into one expression:
6x^2y^3</span><span>√3x</span><span>

</span>
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Tran checked the time on his watch after he finished his daily run select the time that Tran finished running mark all that appl
Viefleur [7K]

Answer:

(a) and (b) are correct

Step-by-step explanation:

Given

See attachment for complete question

Required

Select the correct options

A clock is divided into two equal parts.

(1) 1 - 6

(2) 7 - 12

<em>When the hour hand is on (1), then the time is "after"</em>

<em>When the hour hand is on (2), then the time is "before"</em>

<em />

From the attached clock, we have:

(1) The hour hand on 9

(2) The minute hand 1 unit above 9 i.e. 46 minutes

(1) implies that, the time is before 9 i.e. past 8

(2) implies that the minutes is (60 - 46) before 9 or 46 minutes after 8

In (1) and (2) above, we have two interpretations.

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2: 46 minutes past 8

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3 0
2 years ago
What is the correct answer?
Tasya [4]

Given:

The function is

f(x)=(x+3)^2(x-5)^6

To find:

The zeros of the given function.

Solution:

The general form of polynomial is

P(x)=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}       ...(i)

where, a is a constant, c_1,c_2,...,c_n are zeros of respective multiplicities m_1,m_2,...,m_n.

We have,

f(x)=(x+3)^2(x-5)^6

On comparing this with (i), we get

c_1=-3,m_1=2

c_2=5,m_2=6

It means, -3 is a zero with multiplicity 2 and 5 is a zero with multiplicity 6.

Therefore, the correct option is B.

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2 years ago
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