Rational Expression & Financial Arithmetic

Rational Expression & Financial Arithmetic

6th - 8th Grade

20 Qs

quiz-placeholder

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Rational Expression & Financial Arithmetic

Rational Expression & Financial Arithmetic

Assessment

Quiz

Mathematics

6th - 8th Grade

Practice Problem

Hard

Created by

noureen afshan

Used 1+ times

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20 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 10 pts

Answer explanation

To find the product, multiply the numerators and the denominators: \(\frac{5y^2 \cdot 9x}{3x \cdot 10y} = \frac{45y^2}{30xy} = \frac{3y}{2}\). Thus, the correct answer is \(\frac{3y}{2}\).

2.

MULTIPLE CHOICE QUESTION

1 min • 10 pts

Answer explanation

To find the product, multiply the numerators and denominators: \(\frac{9x^5z^2 \cdot 16y^2}{8y^6 \cdot 3x^3z^2} = \frac{144x^5y^2}{24y^6x^3} = \frac{6x^2}{y^4}\). Thus, the correct answer is \(\frac{6x^2}{y^4}\).

3.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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Answer explanation

To evaluate \( \frac{60x^3}{12x} \) at \( x = -1 \), first simplify: \( \frac{60(-1)^3}{12(-1)} = \frac{60(-1)}{-12} = \frac{60}{12} = 5 \). Thus, the answer is \( 5 \).

4.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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Answer explanation

Substituting x = -7 into the expression gives \frac{5(-7-2)}{5(-7)-10} = \frac{5(-9)}{-35} = \frac{-45}{-35} = \frac{9}{7} = 1. Thus, the correct answer is 1.

5.

MULTIPLE CHOICE QUESTION

15 mins • 1 pt

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Answer explanation

Substituting x=2 into the expression \(\frac{x+4}{x^2+6x+8}\) gives \(\frac{2+4}{2^2+6\cdot2+8} = \frac{6}{4+12+8} = \frac{6}{24} = \frac{1}{4}\). Thus, the correct answer is \(\frac{1}{4}\).

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

When adding rational expressions, you must add numerators together and denominators together.

True

False

Answer explanation

The statement is false. When adding rational expressions, you must find a common denominator first, then add the numerators together while keeping the common denominator unchanged.

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

None

Answer explanation

The rational expression is undefined when the denominator is zero. Setting \( (x-4)(x+3) = 0 \) gives \( x=4 \) and \( x=-3 \). Thus, the correct answer is \( x=4, x=-3 \).

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