Maura and Latasha wrote these steps to find the sum of 1 and One-fifth. Maura's work. 1 and 2-thirds plus 1-fifth. 6-thirds plus 1-fifth. 30-fifteenths plus 3-fifteenths. 33-fifteenths equals 2 and 3-fifteenths. Latasha's work. 1 and 2-thirds plus 1-fifth. (1 plus 0) plus (2-thirds plus 1-fifth). (1) plus (10-fifteenths plus 3-fifteenths). 1 and 13 fifteenths. Which best describes the correct sum? Maura is correct because mixed numbers must be converted to improper fractions to add. Maura is correct because StartFraction 30 over 15 EndFraction is equivalent to 1Two-thirds. Latasha is correct because StartFraction 10 over 15 EndFraction is equivalent to . Latasha is correct because the sum of a mixed number and a fraction is always a mixed number.

Latasha is correct because the sum of a mixed number and a fraction is always a mixed number.

Step-by-step explanation:

Maura and Latasha wrote these steps to find the sum of 1 2 - thirds and One-fifth. Maura's work. 1 and 2-thirds plus 1-fifth. 6-thirds plus 1-fifth. 30-fifteenths plus 3-fifteenths. 33-fifteenths equals 2 and 3-fifteenths. Latasha's work. 1 and 2-thirds plus 1-fifth. (1 plus 0) plus (2-thirds plus 1-fifth). (1) plus (10-fifteenths plus 3-fifteenths). 1 and 13 fifteenths. Which best describes the correct sum?

Solving for the correct sum

The sum of 1 2 - thirds and One-fifth

Expressing this Mathematically

1 2/3 + 1/5

1 + ( 2/3 + 1/5)

The Lowest Common Denominator is 15

1 + (10 + 3/15)

1 13/15

Latasha is correct

Latasha is correct because the sum of a mixed number and a fraction is always a mixed number.

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step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: