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Tutor profile: Holli D.

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Holli D.
High School Math Teacher for Seven Years
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Questions

Subject: Geometry

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Question:

Given an isosceles triangle ABC, if sides AB and BC are congruent, and $$\angle A$$ is 40 degrees, what is the degree measure of $$\angle B$$ ?

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Holli D.
Answer:

Since the side lengths of AB and BC are congruent, that means that angle B is in the center of the isosceles triangle, and therefore angle A and angle C are congruent angles. If angle A and C are both 40 degrees, that takes up 80 degrees in the triangle when added together. You will need to remember that all triangles have a total of 180 degrees. You can then subtract the 80 degrees from the total of 180, leaving angle B to have a measure of 100 degrees.

Subject: Pre-Algebra

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Question:

What is the product of $$(2x^{4})^{3}$$ and $$(5x^{2})$$

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Holli D.
Answer:

When multiplying monomials, the coefficients can be simplified in to one number by multiplying those coefficients, and any bases or variables that are the same can be simplified together also. For this problem you would first want to simplify the first expression. The exponent of 3 belongs to both the base of 2, and the base of x to the power of 4. When you simplify $$2^{3}$$, it is equal to 2(2)(2), which is 8. When you simplify $$(x^{4})^{3}$$, that becomes x(x)(x)(x) to the power of three... therefore becoming x(x)(x)(x) times x(x)(x)(x) times x(x)(x)(x) , equaling $$x^{12}$$. You will then be multiplying the coefficient 8 times the coefficient 5 from the second term. And multiplying $$x^{12}$$ by $$x^{2}$$. 8 times 5 equals 40. Since the exponents tell you how many times to multiply the base by itself, then $$x^{12}$$ times $$x^{2}$$ in expanded form is the same as (x)(x)(x)(x)(x)(x)(x)(x)(x)(x)(x)(x)(x)(x), which is the same as $$x^{14}$$. (A 'shortcut' or rule you can follow when multiplying the same base with exponents is to leave the base the same and add the exponents. So 12 + 2 would also give you the new exponent of 14.) All of this results in your final answer as $$40x^{14}$$.

Subject: Algebra

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Question:

Suppose you are collecting tickets at a football game. Reserved seat tickets cost $4 each and general admission tickets costs $3 each. After the game is over, the turnstile count shows that 1787 people paid admission. You count a total of $5792 from the sale of tickets. How many of each kind of ticket were there?

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Holli D.
Answer:

There are two unknowns in this scenario - the number of reserved seat tickets and the number of general admission tickets. These unknown amounts can be represented by two different variables, and we have enough information to write two different equations. Let's use R for the number of reserved tickets, and G for the number of general admission tickets. If there was a total of 1787 tickets sold, then R + G = 1787. We also could say that the total number minus the reserved number, leaves us with the general admission number. So G = 1787 - R. Knowing the price of each ticket and the total amount of money made, we know 4R + 3G = 5792. Since we know what "G" equals, that amount can then replace the "G" variable in our second equation. Instead of multiplying the $3 by our variable G, we know that 4R + 3(1787 - R) = 5792. We now have a one-variable equation that can be solved to find the number of reserved tickets, "R". This is known as the substitution method for solving systems of equations. Using the equation above, we would multiply the 3 by both terms in parentheses giving us 4R + 5361 - 3R = 5792. Then when we simplify the left side of the equation we know that 1R + 5361 = 5792. This means that some number, R, added to 5361 is equal to 5792. To find this missing variable R, we would take away 5361 from 5792, leaving us with R = 431. This means that 431 reserved seat tickets were sold. To find the number of general admission seats sold we can then take away the "R" amount from the total number of tickets. 1787 - 431 = 1356, telling us that 431 reserved seat tickets were sold and 1356 general admission tickets were sold.

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