Question

iii. challenge problems

1. what is the area of a rectangle with length 7xy² inches and width (8x²y) inches? write your answer as an expression in terms of x and y.
2. use $\frac{x^{2}}{x^{2}}$ and the rules of dividing monomials to explain why $x^{0}=1$.

$\frac{x^{2}}{x^{2}} = 1$ because $2 - 2=0$ and when you subtract with the same number it is 1.

1. correct the error

there is an error in the student work shown below:
question: simplify $\frac{80x^{9}}{16x^{4}}$.
solution:
$\frac{80x^{9}}{16x^{4}}
=\frac{64x^{9}}{x^{4}}
=64x^{5}$
what is the error? explain how to solve the problem.

Explanation:

23.

Step1: Recall area formula

The area $A$ of a rectangle is $A = \text{length}\times\text{width}$.

Step2: Substitute given values

Here, length $l = 7xy^{2}$ and width $w=8x^{2}y$. So $A=(7xy^{2})\times(8x^{2}y)$.

Step3: Multiply coefficients and variables

Multiply the coefficients $7\times8 = 56$, and for the variables, use the rule $a^{m}\times a^{n}=a^{m + n}$. So $x\times x^{2}=x^{1 + 2}=x^{3}$ and $y^{2}\times y=y^{2+1}=y^{3}$. Then $A = 56x^{3}y^{3}$.

Step1: Apply division - rule of monomials

The rule for dividing monomials $\frac{a^{m}}{a^{n}}=a^{m - n}$. For $\frac{x^{2}}{x^{2}}$, where $a = x$, $m = 2$ and $n = 2$.

Step2: Calculate the exponent

Using the rule, $\frac{x^{2}}{x^{2}}=x^{2-2}=x^{0}$. And since any non - zero number divided by itself is 1, $\frac{x^{2}}{x^{2}} = 1$. So $x^{0}=1$ (for $x
eq0$).

Step1: Identify the error

The error is in the first step of the student's work. When dividing $\frac{80x^{9}}{16x^{4}}$, the student subtracted 16 from 80 instead of dividing 80 by 16.

Step2: Correct division of coefficients and variables

Divide the coefficients: $\frac{80}{16}=5$. Divide the variables using the rule $\frac{a^{m}}{a^{n}}=a^{m - n}$, so $\frac{x^{9}}{x^{4}}=x^{9 - 4}=x^{5}$. Then $\frac{80x^{9}}{16x^{4}}=5x^{5}$.

Answer:

$56x^{3}y^{3}$

24.