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Yuliya22 [10]
3 years ago
15

Show me how you solve it

Mathematics
1 answer:
julia-pushkina [17]3 years ago
7 0

Answer:

5^20

Step-by-step explanation:

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u>

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n} }

Therefore:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  }

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{( {a}^{m} ) ^{n} =  {a}^{m \times n}  }

Thus:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  = {5}^{5 \times 4}   } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =   {5}^{20} }

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Using a linear approximation, estimate f(2.1), given that f(2) = 5 and f'(x) = √3x-1.
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Answer:

f\left( {2.1} \right) \approx 5.22360.

Step-by-step explanation:

The linear approximation is given by the equation

                            {f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).}

Linear approximation is a good way to approximate values of f(x) as long as you stay close to the point x= a, but the farther you get from x=a, the worse your approximation.

We know that,

a=2\\f(2) = 5\\f'(x) = \sqrt{3x-1}

Next, we need to plug in the known values and calculate the value of f(2.1):

{L\left( x \right) = f\left( 2 \right) + f^\prime\left( 2 \right)\left( {x - 2} \right) }=5+\sqrt{3(2)-1}(x-2) =5+\sqrt{5}(x-2)

Then

f\left( {2.1} \right) \approx 5+\sqrt{5}(2.1-2)\approx5.22360.

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