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Tutor profile: Paula S.

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Paula S.
Elementary Education graduate with a minor in mathematics
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Questions

Subject:Pre-Algebra

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Question:

Solve and express your answer in simplest form: $$\frac{7}{9}$$+$$\frac{1}{4}$$=

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Paula S.

To add fractions with different bases, we first have to find a common denominator between the 2 fractions. This can be done by finding the Least Common Multiple of 9 and 4. In this case, the least common multiple is simply the product of the two numbers. 36. We can convert $$\frac{7}{9}$$ into a fraction with denominator 36 by multiplying it with the fraction $$\frac{4}{4}$$. This does not change the value of the fraction, because $$\frac{4}{4}=1$$. We can similarly multiply $$\frac{1}{4}$$ by $$\frac{9}{9}$$. Thus the problem is now $$\frac{28}{36}$$+$$\frac{9}{36}$$. Now we can add the numerators and retain the denominator of 36. Therefore $$\frac{28}{36}$$+$$\frac{9}{36}$$=$$\frac{37}{36}$$. However, this is not the final answer because our fraction must be expressed in simplest form. $$\frac{37}{36}$$ is an improper fraction because the numerator is higher than the denominator, meaning the fraction has a value greater than 1. To put this in simplest form, we must find the highest multiple of 36 that can divide into the numerator (in this case 36) and divide it into the numerator. The resulting divisor ($$1$$) becomes our whole number, while the remainder ($$1$$) is our new numerator. The denominator does not change. Thus $$\frac{37}{36}$$ becomes 1$$\frac{1}{36}$$

Subject:Basic Math

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Question:

$$713+372=$$

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Paula S.

To add these numbers, we begin in the ones place. $$3+2=5$$. Because this sum is less than 10, we move on to the tens place. $$1+7=8$$. Again we can move on to the hundreds place. $$7+3=10$$. We keep the 0 from the 10 in the hundreds place and move the 1 into the thousands place. Therefore, $$713+372=1085$$

Subject:Algebra

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Question:

Jennifer was selling desserts at a bake sale. She sold 7 more bags of fudge than cakes and twice as many cookies as cakes. She charged $7 for cakes,$5 for fudge and $3 for cookies. If Jennifer made$215, how many of each item did she sell?

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Paula S.

First, we write expressions for the number of cookies and bags of fudge sold. Let x be the number of cakes sold. Then $$x+7$$ is the amount of fudge sold, and $$2x$$ is the number of cookies sold. Sincce cakes cost $7, fudge costs$5 and cookies cost \$3, we can write an equation for the amount of money Jennifer made. Using the developed expressions, we can write the equation $$7(x) + 5(x+7) + 3(2x)= 215$$. Following the order of operations allows us to solve in this way: $$7(x)+(5x +35) +6x= 215$$ (Distributive Property) $$7x+5x+6x+35=215$$ (Associative Property) $$18x+35=215$$ (Combining Like Terms) $$18x=180$$ (Properties of Equality for Addition and Subtraction) $$x=10$$ (Properties of Equality for Multiplication and Division) We can then plug this value of $$x$$ into the initial expressions to determine the amount of fudge and cookies sold. Therefore, Jennifer sold 10 cakes, 17 bags of fudge and 20 cookies at her bake sale.

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