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zhuklara [117]
4 years ago
11

Simplify the following expression

Mathematics
2 answers:
saveliy_v [14]4 years ago
6 0
Answer:
2√6

Explanation:
Before we begin, remember the following:
√a * √b = √ab
√a * √a = a

Now, for the given:
\frac{6 \sqrt{2} }{ \sqrt{3} }

Multiply both the numerator and the denominator by √3:
\frac{6 \sqrt{2} }{ \sqrt{3} } * \frac{ \sqrt{3}}{ \sqrt{3} }

= \frac{6  \sqrt{2}   \sqrt{3} }{ \sqrt{3}   \sqrt{3} }

= \frac{6 \sqrt{6} }{3}

This can be rewritten as:
\frac{6}{3}  \sqrt{6} = 2√6

Hope this helps :)
LenKa [72]4 years ago
3 0
We have the following expression:
 (6root (2)) / (root (3))
 Rationalizing we have:
 (6root (2)) / (root (3)) * (root (3) / root (3))
 Rewriting we have:
 (6root (6)) / (3)
 2root (6)
 Answer:
 The simplified expression for this case is given by:
 2root (6)
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An element with mass 210 grams decays by 8.3% per minute. How much of the element is remaining after 15 minutes, to the nearest
GarryVolchara [31]

Answer:

57.2\ g

Step-by-step explanation:

In this problem we have a exponential function of the form

f(x)=a(b^x)

where

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x is the time in minutes

a is the initial value (y-intercept of the exponential function)

b is the base

r is the rate

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In this problem we have

a=210\ g

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b=(1-0.083)=0.917

substitute

The exponential function is equal to

f(x)=210(0.917^x)

For x=15 minutes

substitute in the function

f(x)=210(0.917^15)

f(x)=57.2\ g

4 0
3 years ago
Which expression is equivalent to 7a³ (8a² + a)²-4a³?<br> Simplify.
valentina_108 [34]

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2 years ago
T plus 8 times 18 equals 108
Luda [366]

Answer:

T= -36

Step-by-step explanation:

T + 8 x 18 = 108

Multiply 8 and 18

T + 144 = 108

Subtract 144 from both sides to get T alone

T = -36

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Mrrafil [7]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find Y. round to the nearest tenth.
masya89 [10]

9514 1404 393

Answer:

  32.7°

Step-by-step explanation:

Solve the given equation for C, then fill in the given values and evaluate.

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__

Y is angle A in the attached triangle solver.

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