3 because it is a better Dona a toon
The average decrease in value can be solve by first solving the value at year 1 and year 2
and i think the equation is <span>f(x)=10,000(0.73)^x</span>
at x = 1
<span>f(x) = 10,000(0.73)^x
</span><span>f(1) = 10,000(0.73)^(1)
f(1) = 7300
at x = 2
</span><span>f(x) = 10,000(0.73)^x
</span><span>f(2) = 10,000(0.73)^(2)
f(2) = 5329
so the average decrease = ( 7300 - 5329) = $ 1971 per year</span><span>
</span>
To compare fractions with unlike denominators convert them to equivalent fractions with the same denominator. Compare fractions: If denominators are the same you can compare the numerators. The fraction with the bigger numerator is the larger fraction.
