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Aliun [14]
2 years ago
9

Jeremy likes to paint. he estimates the number of paintings he completes using the function p of w equals one third times w plus

four, where w is the number of weeks he spends painting. the function j(y) represents how many weeks per year he spends painting. which composite function would represent how many paintings jeremy completes in a year? p of j of y equals j times the quantity one third times w plus four j of p of w equals one third times j of y plus four p of j of y equals one third times j of y plus four j of p of w equals j times the quantity one third times w plus four
Mathematics
1 answer:
arlik [135]2 years ago
6 0

composite function J[P(W)=J(1/3w+4) represent paintings Jeremy completes in a year .

this equation means number of paintings= weeks(rate)

The function P takes a number of weeks as an argument and returns the number of paintings.

The function J takes some argument (unspecified) and returns a number of weeks per year.

The composite function that will give the number of paintings per year will be P(weeks per year) = P(J(y)).  

P(J(y)) = 1/3·J(y) +4 .

Looking at the units of the input and output of each of the functions is called "units analysis."

<h3>What is Unit analysis?</h3>

Unit analysis means using the rules of multiplying and reducing fractions to solve problems involving different units.

To learn more about unit analysis from the given link

brainly.com/question/14742503

#SPJ4

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a carpet store charges 150 to install 12 square yards of carpet how much would it be for 24 square yards
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Step-by-step explanation:

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2 years ago
How many positive integers $n$ from 1 to 5000 satisfy the congruence $n \equiv 5 \pmod{12}$?
irga5000 [103]
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means that n-5 is a multiple of 12.

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n-5=12k, for some integer k

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n=12k+5


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for k= 0, n=0+5=5 (the first positive integer n, is for k=0)


we solve 5000=12k+5 to find the last k

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k=4995/12=416.25

so check k = 415, 416, 417 to be sure we have the right k:

n=12k+5=12*415+5=4985

n=12k+5=12*416+5=4997

n=12k+5=12*417+5=5009


The last k which produces n<5000 is 416


For all k∈{0, 1, 2, 3, ....416}, n is a positive integer from 1 to 5000,

thus there are 417 integers n satisfying the congruence.


Answer: 417

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You can use the qudratic formula but this will factor to

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Two solutions

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Graphically, (green line is 112 ft)


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