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trasher [3.6K]
3 years ago
14

Please help!! CORRECT ANSWERS ONLY! Please show your work!

Mathematics
2 answers:
statuscvo [17]3 years ago
4 0

Answer:

Y=-9

W=9

Step-by-step explanation:

2y/6+3=0

-3=-3

6/1 *2y/6=-3*6/1

2y=-18

2y/2=-18/2

Y=-9

The rest of the work is in the pic

max2010maxim [7]3 years ago
3 0

GIVEN:

  • 2y/6 + 3 = 0 ...(1)
  • 2/3w + 9 = 15 ...(2)

ANSWER:

On solving equation(1), we get

  • 2y/6 = - 3
  • 2y = - 18
  • y = - 9.

Now, solving equation(2), we get

  • 2/3w + 9 = 15
  • 2/3w = 15 - 9
  • 2/3w = 6
  • w = 6 × 3/2
  • w = 9.

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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 )

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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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