Y=f(x)=(1/2)^x I need the value of f(-7) and the value of f(5) rounded to the nearest ten thousandth.
The value of f(-7) is .( 0.0078) (0.78) (64) (128) The value of f(5), rounded to the nearest ten thousandth, is . (0.0309) (.0312) (0.313) (32)

Respuesta :

Answer:

f(-7)=128, f(5)=0.0312

Step-by-step explanation:

f(x) is the same thing as y, so correctly written its: f(x)=(1/2)^x

plug -7 in for x: f(-7)=(1/2)^(-7)

simplify by doing (1/2) or .5 raised to the power of -7

f(-7)=128

f(5)=(1/2)^5

(.5)^5

=0.0312

Answer:

1) Option 4 - 128

2) Option 2 - 0.0312

Step-by-step explanation:

Given : Expression [tex]Y=f(x)=(\frac{1}{2})^x[/tex]

To find : The value of f(-7) and the value of f(5) rounded to the nearest ten thousandth?

Solution :

The given expression is [tex]Y=f(x)=(\frac{1}{2})^x[/tex]

1) Substitute x=-7

[tex]f(-7)=(\frac{1}{2})^{-7}[/tex]

As [tex](\frac{1}{a})^{-n}=a^n[/tex]

[tex]f(-7)=2^7[/tex]

[tex]f(-7)=128[/tex]

Therefore, Option 4 is correct.

2) Substitute x=5

[tex]f(5)=(\frac{1}{2})^{5}[/tex]

[tex]f(5)=\frac{1}{2^{5}}[/tex]

[tex]f(5)=\frac{1}{32}[/tex]

[tex]f(5)=0.03125[/tex]

[tex]f(5)\approx 0.0312[/tex]

Therefore, Option 2 is correct.

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