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padilas [110]
3 years ago
7

the absolute value function, f(x) = –|x| – 3, is shown. what is the range of the function? all real numbers all real numbers les

s than or equal to 0 all real numbers greater than or equal to –3 all real numbers less than or equal to –3
Mathematics
2 answers:
pickupchik [31]3 years ago
4 0
The range is all real numbers less than or equal to –3
BaLLatris [955]3 years ago
4 0

Answer:

Range is : all real numbers less than or equal to –3

Step-by-step explanation:

Range of a function f(x) is defined as set of all real possible values of f(x) such that the function is defined.

the given function is : -|x| - 3

Now, since absolute function only writes positive value so least value of |x| is 0

⇒ Greatest value of -|x| - 3 is -3

And least value is not fixed So the least possible value of f(x) is -∞

So, the range of f(x) is (-∞,-3]

Hence, Range is : all real numbers less than or equal to –3

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Irina18 [472]

Answer:

\bf C)\: -\cfrac{1}{32}

Step-by-step explanation:

\tt \cfrac{1}{2^{-2}x^{-3}y^5}

\tt x=2\\y=-4

~

Substitute ''x'' with '2' & 'y' with "-4":-

\tt \cfrac{1}{2^{-2}\times \:2^{-3}\left(-4\right)^5}

First let's solve \tt 2^{-2}\times \:\:2^{-3}\left(-4\right)^5

\tt 2^{-2}\times \:2^{-3}=\boxed{\cfrac{1}{32} }

\tt \cfrac{1}{32}\left(-4\right)^5

Calculate exponents:- -4^5= -1024

\tt \cfrac{1}{32}\left(-1024\right)

Remove parentheses, apply rule:- \tt a\left(-b\right)=-ab

\tt-\frac{1}{32}\times \:1024

\tt \cfrac{1}{32}\times \cfrac{1024}{1}

Apply fraction rule:-

\tt -\cfrac{1\times \:1024}{32\times \:1}

\tt \cfrac{1024}{32}

Divide numbers:-

\tt -32

Now that we're done factoring the denominator let's bring down the numerator which is ' 1 ':-

\tt -\cfrac{1}{32}\;\; \bf Answer

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Answer:

Step-by-step explanation:

It’s C

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