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Fantom [35]
2 years ago
12

Use complete sentences to explain in detail how you would use the substitution method to solve the following system of equations

:
x = 4y

x + 2y = 12

Your description should include:

-how to find x

-how to find y

-the answer to the system
Mathematics
1 answer:
-BARSIC- [3]2 years ago
5 0
If we now that 4y equals to x, that means we can replace x with 4y.

Therefore we can write: 4y+2y = 12
We can summarize that to 6y = 12.
If we divide both sides with 6, you get y = 2

In order to solve for x, we can use the second equation and just replace y with value we just got (y=2).

We can write: x+2*2=12
Now you move 2*2 (which is 4) to the other side: x=12-4
Now you get: x = 8
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Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
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Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

To find the simplified form of the given expression :

[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2

=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2 ( using the property (ab)^m=a^m.b^m )

=(2^6)(3^{-12})\times (2^{-6})(3^4) ( using the property (a^m)^n=a^{mn}

=(2^6)(2^{-6})(3^{-12})(3^4) ( combining the like powers )

=2^{6-6}3^{-12+4} ( using the property a^m.a^n=a^{m+n} )

=2^03^{-8}

=\frac{1}{3^8} ( using the property a^{-m}=\frac{1}{a^m} )

Therefore [(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is \frac{1}{3^8} is correct answer

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3 years ago
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