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3 years ago

Can someone please simplify the complex fraction?

Mathematics

1 answer:

dmitriy555 [2]3 years ago

8 0

$\bf \cfrac{\qquad \frac{x+3}{4x^2-16}\qquad }{\frac{2x^2+10x+12}{2x-4}}\implies \cfrac{\frac{x+3}{4(x^2-4)}}{\frac{\underline{2}(x^2+5x+6)}{\underline{2}(x-2)}}\implies \cfrac{\frac{x+3}{4(x^2-2^2)}}{\frac{x^2+5x+6}{x-2}}\implies \cfrac{\frac{x+3}{4(x-2)(x+2)}}{\frac{(x+3)(x+2)}{x-2}}$

$\bf \cfrac{\underline{x+3}}{4\underline{(x-2)}(x+2)}\cdot \cfrac{\underline{x-2}}{\underline{(x+3)}(x+2)}\implies \cfrac{1}{4(x+2)}\cdot \cfrac{1}{x+2}
\\\\\\
\cfrac{1}{4(x+2)(x+2)}\implies \cfrac{1}{4(x^2+4x+4)}\implies \cfrac{1}{4x^2+16x+16}$

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First make the denominators of the fractions the same by multiplying.

$\frac{2}{3}$ × $\frac{2}{2}$
= $\frac{4}{6}$

Now we just need to add the fractions together

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= $\frac{9}{6}$

9/6 would be the answer, but it can be simplified further, then converted into decimal form.

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A jet travels from chicago to kansas city at a speed of 440 knots. If it takes a plane 3 hr longer to fly from kansas city to ch

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d = distance between the two cities

v₁ = average speed while going from chicago to kansas city = 440 knots

t₁ = time taken to travel distance going from chicago to kansas city

time taken to travel distance going from chicago to kansas city is given as

t₁ = d/v₁

t₁ = d/440 eq-1

v₂ = average speed while going from kansas city to chicago = 110 knots

t₂ = time taken to travel distance going from kansas city to chicago

time taken to travel distance going from kansas city to chicago is given as

t₂ = d/v₂

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Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

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