Question

practice distributed and integrated

1. write each decimal as a fraction in lowest terms.
2. 0.24 2. 0.82 3. 0.32
3. what is the measure of an interior angle of a regular pentagon?
4. determine whether (y^{2}=8x) is a function. if so, state the domain and range.
5. find the sum (2\begin{bmatrix}1&0\\0&1end{bmatrix}+\begin{bmatrix}-3&-2\\-4&0end{bmatrix})

*7. if (m) varies directly with (n), and (m = 8) when (n = 2), what is the value of the constant of variation?
graphing calculator find each product. check using the graphing calculator.
8. (\begin{bmatrix}2&-1\\3&0\\1&4end{bmatrix}\times\begin{bmatrix}1&1&-3\\6&-5&0end{bmatrix}) 9. (\begin{bmatrix}-1&-1&-1\\2&5&3\\-5&4&-2end{bmatrix}\times\begin{bmatrix}0\\1\\0end{bmatrix})

Explanation:

- ## Step1: Write 0.24 as a fraction
- 0.24 can be written as $\frac{24}{100}$ since 0.24 means 24 hundredths.
- ## Step2: Simplify the fraction
- Find the greatest common divisor (GCD) of 24 and 100. The GCD of 24 and 100 is 4. Divide both the numerator and the denominator by 4: $\frac{24\div4}{100\div4}=\frac{6}{25}$.
- # Answer: $\frac{6}{25}$
2. **For writing 0.82 as a fraction in lowest - terms**:
- # Explanation:
- ## Step1: Write 0.82 as a fraction
- 0.82 can be written as $\frac{82}{100}$ since 0.82 means 82 hundredths.
- ## Step2: Simplify the fraction
- The GCD of 82 and 100 is 2. Divide both the numerator and the denominator by 2: $\frac{82\div2}{100\div2}=\frac{41}{50}$.
- # Answer: $\frac{41}{50}$
3. **For writing 0.32 as a fraction in lowest - terms**:
- # Explanation:
- ## Step1: Write 0.32 as a fraction
- 0.32 can be written as $\frac{32}{100}$ since 0.32 means 32 hundredths.
- ## Step2: Simplify the fraction
- The GCD of 32 and 100 is 4. Divide both the numerator and the denominator by 4: $\frac{32\div4}{100\div4}=\frac{8}{25}$.
- # Answer: $\frac{8}{25}$
4. **For the measure of an interior angle of a regular pentagon**:
- # Explanation:
- ## Step1: Use the formula for the measure of an interior angle of a regular polygon
- The formula is $\theta=\frac{(n - 2)\times180^{\circ}}{n}$, where $n$ is the number of sides of the polygon. For a pentagon, $n = 5$.
- ## Step2: Calculate the value
- $\theta=\frac{(5 - 2)\times180^{\circ}}{5}=\frac{3\times180^{\circ}}{5}=108^{\circ}$.
- # Answer: $108^{\circ}$
5. **For determining whether $y^{2}=8x$ is a function**:
- # Explanation:
- ## Step1: Recall the vertical - line test for functions
- A relation $y = f(x)$ is a function if for every $x$ - value in the domain, there is exactly one $y$ - value. For the equation $y^{2}=8x$, we can solve for $y$: $y=\pm\sqrt{8x}$.
- ## Step2: Apply the vertical - line test
- If we consider a vertical line $x = a>0$, it will intersect the graph of $y^{2}=8x$ at two points $y=\sqrt{8a}$ and $y =-\sqrt{8a}$. So, it is not a function.
- # Answer: No
6. **For finding the sum of matrices $\begin{bmatrix}1&0\\0&1\end{bmatrix}+\begin{bmatrix}- 3&-2\\-4&0\end{bmatrix}$**:
- # Explanation:
- ## Step1: Recall the rule for matrix addition
- To add two matrices of the same size, we add the corresponding elements. Let $A=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $B=\begin{bmatrix}-3&-2\\-4&0\end{bmatrix}$.
- ## Step2: Perform the addition
- $A + B=\begin{bmatrix}1+( - 3)&0+( - 2)\\0+( - 4)&1 + 0\end{bmatrix}=\begin{bmatrix}-2&-2\\-4&1\end{bmatrix}$.
- # Answer: $\begin{bmatrix}-2&-2\\-4&1\end{bmatrix}$
7. **If $m$ varies directly with $n$ and $m = 8$ when $n = 2$, find the constant of variation**:
- # Explanation:
- ## Step1: Recall the direct - variation formula
- The formula for direct variation is $m=kn$, where $k$ is the constant of variation.
- ## Step2: Solve for $k$
- Substitute $m = 8$ and $n = 2$ into the formula: $8=k\times2$. Divide both sides by 2 to get $k = 4$.
- # Answer: 4
8. **For finding the product $\begin{bmatrix}2&-1\\3&0\\1&4\end{bmatrix}\times\begin{bmatrix}1&1&-3\\6&-5&0\end{bmatrix}$**:
- # Explanation:
- ## Step1: Recall the rule for matrix multiplication
- If $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix. The $(i,j)$ - entry of $AB$ is the dot - product of the $i$ - th row of $A$ and the $j$ - th column of $B$.
- ## Step2: Calculate the product
- The first row of the product:
- $(2\times1+( - 1)\times6,2\times1+( - 1)\times( - 5),2\times( - 3)+( - 1)\times0)=(2 - 6,2 + 5,-6+0)=(-4,7,-6)$.
- The second row of the product:
- $(3\times1+0\times6,3\times1+0\times( - 5),3\times( - 3)+0\times0)=(3,3,-9)$.
- The third row of the product:
- $(1\times1+4\times6,1\times1+4\times( - 5),1\times( - 3)+4\times0)=(1 + 24,1-20,-3)=(25,-19,-3)$.
- So the product is $\begin{bmatrix}-4&7&-6\\3&3&-9\\25&-19&-3\end{bmatrix}$.
- # Answer: $\begin{bmatrix}-4&7&-6\\3&3&-9\\25&-19&-3\end{bmatrix}$
9. **For finding the product $\begin{bmatrix}-1&-1&-1\\2&5&3\\-5&4&-2\end{bmatrix}\times\begin{bmatrix}0\\1\\0\end{bmatrix}$**:
- # Explanation:
- ## Step1: Recall the rule for matrix - vector multiplication
- If $A$ is an $m\times n$ matrix and $\mathbf{x}$ is an $n\times1$ vector, then $A\mathbf{x}$ is an $m\times1$ vector. The $i$ - th entry of $A\mathbf{x}$ is the dot - product of the $i$ - th row of $A$ and the vector $\mathbf{x}$.
- ## Step2: Calculate the product
- The first row: $(-1)\times0+( - 1)\times1+( - 1)\times0=-1$.
- The second row: $2\times0 + 5\times1+3\times0=5$.
- The third row: $(-5)\times0+4\times1+( - 2)\times0=4$.
- So the product is $\begin{bmatrix}-1\\5\\4\end{bmatrix}$.
- # Answer: $\begin{bmatrix}-1\\5\\4\end{bmatrix}$

Answer:

1. For writing 0.24 as a fraction in lowest - terms:

• # Explanation:
• ## Step1: Write 0.24 as a fraction
• 0.24 can be written as $\frac{24}{100}$ since 0.24 means 24 hundredths.
• ## Step2: Simplify the fraction
• Find the greatest common divisor (GCD) of 24 and 100. The GCD of 24 and 100 is 4. Divide both the numerator and the denominator by 4: $\frac{24\div4}{100\div4}=\frac{6}{25}$.
• # Answer: $\frac{6}{25}$

1. For writing 0.82 as a fraction in lowest - terms:

• # Explanation:
• ## Step1: Write 0.82 as a fraction
• 0.82 can be written as $\frac{82}{100}$ since 0.82 means 82 hundredths.
• ## Step2: Simplify the fraction
• The GCD of 82 and 100 is 2. Divide both the numerator and the denominator by 2: $\frac{82\div2}{100\div2}=\frac{41}{50}$.
• # Answer: $\frac{41}{50}$

1. For writing 0.32 as a fraction in lowest - terms:

• # Explanation:
• ## Step1: Write 0.32 as a fraction
• 0.32 can be written as $\frac{32}{100}$ since 0.32 means 32 hundredths.
• ## Step2: Simplify the fraction
• The GCD of 32 and 100 is 4. Divide both the numerator and the denominator by 4: $\frac{32\div4}{100\div4}=\frac{8}{25}$.
• # Answer: $\frac{8}{25}$

1. For the measure of an interior angle of a regular pentagon:

• # Explanation:
• ## Step1: Use the formula for the measure of an interior angle of a regular polygon
• The formula is $\theta=\frac{(n - 2)\times180^{\circ}}{n}$, where $n$ is the number of sides of the polygon. For a pentagon, $n = 5$.
• ## Step2: Calculate the value
• $\theta=\frac{(5 - 2)\times180^{\circ}}{5}=\frac{3\times180^{\circ}}{5}=108^{\circ}$.
• # Answer: $108^{\circ}$

1. For determining whether $y^{2}=8x$ is a function:

• # Explanation:
• ## Step1: Recall the vertical - line test for functions
• A relation $y = f(x)$ is a function if for every $x$ - value in the domain, there is exactly one $y$ - value. For the equation $y^{2}=8x$, we can solve for $y$: $y=\pm\sqrt{8x}$.
• ## Step2: Apply the vertical - line test
• If we consider a vertical line $x = a>0$, it will intersect the graph of $y^{2}=8x$ at two points $y=\sqrt{8a}$ and $y =-\sqrt{8a}$. So, it is not a function.
• # Answer: No

1. **For finding the sum of matrices

$$\begin{bmatrix}1&0\\0&1\end{bmatrix}+\begin{bmatrix}- 3&-2\\-4&0\end{bmatrix}$$

**:

• # Explanation:
• ## Step1: Recall the rule for matrix addition
• To add two matrices of the same size, we add the corresponding elements. Let $A=

$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

$ and $B=

$$\begin{bmatrix}-3&-2\\-4&0\end{bmatrix}$$

$.

• ## Step2: Perform the addition
• $A + B=

$$\begin{bmatrix}1+( - 3)&0+( - 2)\\0+( - 4)&1 + 0\end{bmatrix}$$

=

$$\begin{bmatrix}-2&-2\\-4&1\end{bmatrix}$$

$.

• # Answer:

$$\begin{bmatrix}-2&-2\\-4&1\end{bmatrix}$$

1. If $m$ varies directly with $n$ and $m = 8$ when $n = 2$, find the constant of variation:

• # Explanation:
• ## Step1: Recall the direct - variation formula
• The formula for direct variation is $m=kn$, where $k$ is the constant of variation.
• ## Step2: Solve for $k$
• Substitute $m = 8$ and $n = 2$ into the formula: $8=k\times2$. Divide both sides by 2 to get $k = 4$.
• # Answer: 4

1. **For finding the product

$$\begin{bmatrix}2&-1\\3&0\\1&4\end{bmatrix}\times\begin{bmatrix}1&1&-3\\6&-5&0\end{bmatrix}$$

**:

• # Explanation:
• ## Step1: Recall the rule for matrix multiplication
• If $A$ is an $m\times n$ matrix and $B$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix. The $(i,j)$ - entry of $AB$ is the dot - product of the $i$ - th row of $A$ and the $j$ - th column of $B$.
• ## Step2: Calculate the product
• The first row of the product:
• $(2\times1+( - 1)\times6,2\times1+( - 1)\times( - 5),2\times( - 3)+( - 1)\times0)=(2 - 6,2 + 5,-6+0)=(-4,7,-6)$.
• The second row of the product:
• $(3\times1+0\times6,3\times1+0\times( - 5),3\times( - 3)+0\times0)=(3,3,-9)$.
• The third row of the product:
• $(1\times1+4\times6,1\times1+4\times( - 5),1\times( - 3)+4\times0)=(1 + 24,1-20,-3)=(25,-19,-3)$.
• So the product is

$$\begin{bmatrix}-4&7&-6\\3&3&-9\\25&-19&-3\end{bmatrix}$$

.

• # Answer:

$$\begin{bmatrix}-4&7&-6\\3&3&-9\\25&-19&-3\end{bmatrix}$$

1. **For finding the product

$$\begin{bmatrix}-1&-1&-1\\2&5&3\\-5&4&-2\end{bmatrix}\times\begin{bmatrix}0\\1\\0\end{bmatrix}$$

**:

• # Explanation:
• ## Step1: Recall the rule for matrix - vector multiplication
• If $A$ is an $m\times n$ matrix and $\mathbf{x}$ is an $n\times1$ vector, then $A\mathbf{x}$ is an $m\times1$ vector. The $i$ - th entry of $A\mathbf{x}$ is the dot - product of the $i$ - th row of $A$ and the vector $\mathbf{x}$.
• ## Step2: Calculate the product
• The first row: $(-1)\times0+( - 1)\times1+( - 1)\times0=-1$.
• The second row: $2\times0 + 5\times1+3\times0=5$.
• The third row: $(-5)\times0+4\times1+( - 2)\times0=4$.
• So the product is

$$\begin{bmatrix}-1\\5\\4\end{bmatrix}$$

.

• # Answer:

$$\begin{bmatrix}-1\\5\\4\end{bmatrix}$$