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ziro4ka [17]
3 years ago
9

Hello, can you please express this without denominators with the work?

Mathematics
1 answer:
Lena [83]3 years ago
5 0
\bf  a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
%  negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 
\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}} \\\\
-----------------------------\\\\
\cfrac{4xy}{mn^5}\implies \cfrac{4xy}{1}\cdot \cfrac{1}{m^1}\cdot \cfrac{1}{n^5}\implies 4xym^{-1}n^{-5}

notice, all you do is, move the factor from the bottom to the top, or from the top to the bottom, and the sign changes, from negative to positive or the other way around, is all there's on that
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enelope determined the solutions of the quadratic function by completing the square. f(x) = 4x2 + 8x + 1 –1 = 4x2 + 8x –1 = 4(x2
Marat540 [252]
The correct question is
<span>Penelope determined the solutions of the quadratic function by completing the square.
f(x) = 4x² + 8x + 1
–1 = 4x² + 8x
–1 = 4(x² + 2x)
–1 + 1 = 4(x² + 2x + 1)
0 = 4(x + 2)²
0 = (x + 2)²
0 = x + 2
–2 = x
What error did Penelope make in her work?

we have that
</span>f(x) = 4x² + 8x + 1
to find the solutions of the quadratic function
let
f(x)=0
 4x² + 8x + 1=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

 (4x² + 8x)=-1

Factor the leading coefficient 

 4*(x² + 2x)=-1

Complete the square Remember to balance the equation by adding the same constants to each side.

 4*(x² + 2x+1)=-1+4 --------> ( added 4 to both sides)

Rewrite as perfect squares

4*(x+1)²=3

(x+1)²=3/4--------> (+/-)[x+1]=√3/2

(+)[x+1]=√3/2---> x1=(√3/2)-1----> x1=(√3-2)/2

(-)[x+1]=√3/2----> x2=(-2-√3)/2

therefore

the answer is

<span>Penelope should have added 4 to both sides instead of adding 1.</span>
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Answer:

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Answer:

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Step-by-step explanation:

This is your answer I am pretty sure this is correct.

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