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sergij07 [2.7K]
3 years ago
6

Classify the solutions of 1 over x plus 4, plus one half, equals 1 over x plus 4 as extraneous or non-extraneous.

Mathematics
2 answers:
lara [203]3 years ago
7 0

Answer:

x=-4 , Extraneous

Step-by-step explanation:

Given : Equation - \frac{1}{x+4}+ \frac{1}{2}=\frac{1}{x+4}

To classify : The given equation as extraneous or non-extraneous ?

Solution :  

An extraneous solution is one that we arrive at that will not work in the equation.

For rational equations, extraneous solutions are ones that will cause the denominator to be 0.

So, first we solve the given equation

\frac{1}{x+4}+ \frac{1}{2}=\frac{1}{x+4}

Taking LCM to LHS

\frac{2+x+4}{2(x+4)}=\frac{1}{x+4}

Cancel (x+4) both side,

\frac{x+6}{2}=1

Cross multiply,

x+6=2

x=-4

Now, if we put x=-4 in the equation the denominator is zero.

Therefore, It is an extraneous solution.

katrin2010 [14]3 years ago
3 0

X=-4 ; extraneous is your answer

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2 years ago
Which is a better deal for a package of soap?
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I get the part A of the question its just the part B that i would really appreciate for someone to explain
Julli [10]

Step-by-step explanation:

650065 = (a² + 1)(c² + 1)

and

650065 = (ac - 1)² + (a + c)² = the sum of 2 squared numbers (ac - 1)² and (a + c)².

we see that 5 and 13 and therefore 5×13=65 are all factor of 650065. and then 650065/65 = 10001.

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and 10001 = (10000 + 1) = (100² + 1)

so, a = 8, c = 100

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8 0
2 years ago
What is the solution to 2 log Subscript 9 Baseline (x) = log Subscript 9 Baseline 8 + log Subscript 9 Baseline (x minus 2) x = n
Kaylis [27]

Answer:

\large\boxed{\large\boxed{x=4}}

Step-by-step explanation:

The equation to solve is:

            2\log_9 x=\log_9 8+\log_9 (x-2)

1. <u>On the left-hand side </u>use: "The product of a constant by a logarithm is equal to the logarithm raised to the constant"

Thus, the left-hand side is:

               2\log_9 x=\log_9 x^2

2. On the <u>right-hand side</u> use "The sum of two logarithms with the same base is the logarithm of the product":

Then, on the right-hand side:

        \log_9 8+\log_9 (x-2)=\log_9 8(x-2)

3. <u>Make them equal</u>:

      \log_9 x^2=\log_9 8(x-2)

4. Since the two functions are the same, <u>make the arguments equal</u>:

     x^2=8(x-2)

5. <u>Solve the equation</u>:

        x^2=8x-16\\\\x^2-8x+16=0\\\\(x-4)^2=0\\\\x-4=0\\\\x=4\leftarrow solution

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3 years ago
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