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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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A comedy show sells lower level tickets for $77 and upper level tickets for $99. On the opening night 2615 tickets were sold for
babunello [35]

Let x represent lower level tickets that cost $77

Let y represent upper level tickets that cost $99

Cost equation: 77x + 99y = 247,071

Tickets equation: x + y = 2615

Using the elimination method, multiply the second equation by -77:

77x + 99y = 247,071

-77x - 77y = -201,355

--> 22y = 45,716

--> y = 2,078

Now plug "y" into either equation and solve for "x". I chose the Tickets equation. x + y = 2615 → x = 2615 - y → x = 2615 - 2078 → x = 537

Answer: lower level = 537 tickets, upper level = 2078 tickets.


7 0
3 years ago
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Answer:

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8 0
2 years ago
A standard letter-sized piece of paper has a length and width of 8.5 inches by 11 inches.
den301095 [7]

Answer:

Area of the resulting rectangle = 23.375 inch²

Step-by-step explanation:

After 1 fold width or length reduces to half.

Area = Length x width

So after 1 fold area will reduce to half.

So we have

              \texttt{Area after n folds =}\frac{\texttt{Initial area}}{2^{\texttt{number of folds}}}

Initial area = 8.5 x 11 = 93.5 inch²

Here the paper were folded completely in half, length is reduced to half and width reduced to half, that is

Number of folds = 2

\texttt{Area after 2 folds =}\frac{\texttt{Initial area}}{2^{2}}\\\\\texttt{Area after 2 folds =}\frac{93.5}{4}=23.375inch^2

Area after 2 folds = 23.375 inch²

Area of the resulting rectangle = 23.375 inch²

8 0
3 years ago
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Answer:

=27a^3b^3+54a^2b^2c+36abc^2+8c^3

Step-by-step explanation:

(3ab+2c)^3

=(3ab+2c) × (3ab+2c) ×

=(3ab+2c)(9a^2b^2+12abc+4c^2)

=(3ab)(9a^2b^2)+(3ab)(12abc)+(3ab)(4c^2)+(2c)(9a^2b^2)+(2c)(12abc)+(2c)(4c^2)

=27a^3b^3+36a^2b^2c+12abc^2+18a^2b^2c+24abc^2+8c^3

=27a^3b^3+54a^2b^2c+36abc^2+8c^3

<em>hope this helps....</em>

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