rational operations and simplifying

rational operations and simplifying

10th - 12th Grade

81 Qs

quiz-placeholder

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rational operations and simplifying

rational operations and simplifying

Assessment

Quiz

Mathematics

10th - 12th Grade

Hard

CCSS
HSA.APR.D.6, HSA.APR.D.7, HSA.APR.A.1

Standards-aligned

Created by

Marilee Wyckoff

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81 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do we find the restrictions of a rational expression?

We set the numerator equal to zero and solve for the variable

We set the denominator equal to zero and solve for the variable

We plug zero into the numerator and simplify

We plug zero into the denominator and simplify

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Write this product in simplest form:
  6a2b15c35c4a\frac{6a^2b}{15c^3}\cdot\frac{5c}{4a}  

 30a2bc60ac3\frac{30a^2bc}{60ac^3}  

 a2bc2ac3\frac{a^2bc}{2ac^3}  

 30a2bc60ac3\frac{30a^2bc}{60ac^3}  

 ab2c2\frac{ab}{2c^2}  

Tags

CCSS.HSA.APR.D.7

3.

MULTIPLE SELECT QUESTION

30 sec • 1 pt

What are the restrictions of:   19u8u(u+2)\frac{19u}{8u\left(u+2\right)}  

(Pick all that apply!)

 u 0u\ \ne0  

 u 2u\ \ne2  

 u 2u\ \ne-2  

 u8u\ne8  

 u8u\ne-8  

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Write the following quotient in simplest form:

 8p8p+3÷p1p+3\frac{8p-8}{p+3}\div\frac{p-1}{p+3}  

 88  

 8p1p+3\frac{8p-1}{p+3}  

 8(p1)(p+3)(p+3)(p1)\frac{8\left(p-1\right)\left(p+3\right)}{\left(p+3\right)\left(p-1\right)}  

 8(p1)(p+3) + (p1)(p+3)(p+3)(p1)\frac{8\left(p-1\right)\left(p+3\right)\ +\ \left(p-1\right)\left(p+3\right)}{\left(p+3\right)\left(p-1\right)}  

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Write the following difference in simplest form:

 aa+32a3\frac{a}{a+3}-\frac{2a}{3}  

 2a2+9a3(a+3)\frac{-2a^2+9a}{3\left(a+3\right)}  

 2a2+9a3(a+3)\frac{2a^2+9a}{3\left(a+3\right)}  

 2a23a3(a+3)\frac{-2a^2-3a}{3\left(a+3\right)}  

 (aa2)3(a+3)\frac{\left(a-a^2\right)}{3\left(a+3\right)}  

Tags

CCSS.HSA.APR.D.7

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Student 1 says that the restrictions of 9(d4)(d7)(d+8)\frac{9\left(d-4\right)}{\left(d-7\right)\left(d+8\right)} are d4d\ne4 , d7d\ne7 , and d8d\ne-8 . Is Student 1 correct? Why?

Yes because d = 4, d = 7, and d = -8 all make the denominator equal to zero

Yes because d = 4, d = 7, and d = -8 all make the expression evaluate to zero

No because d = 4 does not make the denominator equal to zero

No because the real restrictions are d4d\ne-4 , d7d\ne-7 , and d8d\ne8

No because d = 0 is also a restriction

Tags

CCSS.HSA.APR.D.7

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Write this product in simplest form:

 h2+3h24h+3\frac{h^2+3h}{2}\cdot\frac{4}{h+3}  

 4(h2+3h)2(h+3)\frac{4\left(h^2+3h\right)}{2\left(h+3\right)}  

 2h2+3hh+3\frac{2h^2+3h}{h+3}  

 2h2h  

 2(h2+3h)h+3\frac{2\left(h^2+3h\right)}{h+3}  

Tags

CCSS.HSA.APR.D.7

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