1) Number of letters Matilda has sorted after x hours: m(x)Matilda has already sorted 50 letters and continues sorting at a rate of 50 letters per hour:m(x)=50+50xwhere:Number of hours: x
Number of letters Lorraine has sorted after x hours: l(x)Lorraine has already sorted 80 letters and continues sorting at a rate of 40 letters per hour:l(x)=80+40xwhere:Number of hours: x
Which function can Matilda and Lorraine use to determine the total number of letters they have sorted after x hours?Total number of letters they have sorted after x hours: f(x)
f(x)=m(x)+l(x)f(x)=(50+50x)+(80+40x)f(x)=50+50x+80+40xf(x)=90x+130
Answer: The function Matilda and Lorraine can use to determine the total number of letters they have sorted after x hours is f(x)=90x+130
2) How many letters will they have sorted after 6 hours?
x=6→f(6)=?f(6)=90(6)+130f(6)=540+130f(6)=670
Answer: They will have sorted 670 letters after 6 hours
Answer: First option: The function that describes the total number of letters sorted by Matilda and Lorraine in x hours is given by f(x) = 90x + 130. Thus, they will have sorted 670 letters in 6 hours.
The answer is 385mi. 110mi/2hr = 55mi/1hr=385mi/7hr
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Seems to be an even function.
Answer: Choice D
b greater-than 3 and StartFraction 2 over 15 EndFraction
In other words,
b > 3 & 2/15
or
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Explanation:
Let's convert the mixed number 2 & 3/5 into an improper fraction.
We'll use the rule
a & b/c = (a*c + b)/c
In this case, a = 2, b = 3, c = 5
So,
a & b/c = (a*c + b)/c
2 & 3/5 = (2*5 + 3)/5
2 & 3/5 = (10 + 3)/5
2 & 3/5 = 13/5
The inequality 
is the same as 
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Let's multiply both sides by 15 to clear out the fractions
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Now isolate the variable b
Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how
47/15 = 3 remainder 2
The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.
