isaiahrobertson49 isaiahrobertson49
  • 15-12-2021
  • Mathematics
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10. If f(x)=40(.97)* +78 , find the value of x,
to the nearest ten- thousandth, that solves the
equation f(x)=110.

10 If fx4097 78 find the value of x to the nearest ten thousandth that solves the equation fx110 class=

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ThePerfectionOne01
ThePerfectionOne01 ThePerfectionOne01
  • 15-12-2021

Answer:

[tex]f(x)=79[/tex]

Step-by-step explanation:

Given: [tex]f(x)=40(.97)^\bold{x} +78[/tex]

Plug in the function [tex]f(x)=110[/tex] for [tex]x[/tex]; solve:

[tex]f(x)=40(.97)^\bold{x} +78[/tex]

[tex]40(.97)^\bold{110} +78[/tex]

Exponents:

[tex]0.97^{110}=0.03506[/tex]

[tex]=40\cdot \:0.03506 ^\bold{110} +78[/tex]

[tex]=1.40265 + 78[/tex]

[tex]= 79.40265[/tex]

Therefore, [tex]\bold{f(x)=79.40265}[/tex].

79.40256 in the nearest tenths is 79 Learn how to: https://brainly.com/question/192096

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