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katrin2010 [14]
3 years ago
6

4|x+7|+5>17 solve the following

Mathematics
1 answer:
katrin2010 [14]3 years ago
7 0

Answer:

  • 4|x+7|+5>17
  • 4x+28+5>17
  • 4x>17-33
  • x>-16/4
  • x>-4

<h3>hope it helps..</h3><h3>stay safe healthy and happy...</h3>
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Plz help me with this
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Only the last two options are true.

The first one is false, because 4.5 is the value of the maximum of f(x), not the point where it is reached.

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The third one is true, because the maximum value of f(x) is 4.5, and the maximum value of g(x) is 9, which is twice the maximum of f(x)

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3 years ago
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Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega
AlekseyPX

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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=\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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