awakwe / Algebra-2

Algebra 2

Unit 8: EXPONENTS AND RADICALS

Simplify the expressions using exponent rules.

1. $b^4 \cdot b^3$

   Solution: When multiplying expressions with the same base, we add the exponents. So, $b^4 \cdot b^3 = b^{4+3} = b^7$.

2. $c^5 \cdot c^2 \cdot c^2$

   Solution: Similarly, for this expression, we add all the exponents: $c^5 \cdot c^2 \cdot c^2 = c^{5+2+2} = c^9$.

3. $a^{-4} \cdot a^{-3}$

   Solution: The rule of adding exponents applies here as well. Negative exponents indicate that the base is on the denominator of a fraction. However, for this problem, we just add the exponents: $a^{-4} \cdot a^{-3} = a^{-4-3} = a^{-7}$.

4. $x^5 \cdot x^{-4} \cdot x$

   Solution: In this case, we add the exponents, and remember that $x$ is the same as $x^1$. So, $x^5 \cdot x^{-4} \cdot x = x^{5+(-4)+1} = x^2$.

5. $(2x)^2 \cdot (4y)^2$

   Solution: When raising a power to a power, we multiply the exponents. Also, remember that both the coefficient and the variable are raised to the power. So, $(2x)^2 \cdot (4y)^2 = (2^2x^2)(4^2y^2) = 4x^2 \cdot 16y^2 = 64x^2y^2$.

6. $-2gh \cdot (g^3h^5)$

   Solution: In this case, we multiply the like terms. So, $-2gh \cdot (g^3h^5) = -2g^1h^1 \cdot g^3h^5 = -2g^{1+3}h^{1+5} = -2g^4h^6$.

7. $10x^2y^3 \cdot (10xy^8)$

   Solution: Here, we multiply the coefficients and add the exponents of like terms. So, $10x^2y^3 \cdot (10xy^8) = 10 \cdot 10 \cdot x^{2+1}y^{3+8} = 100x^3y^{11}$.

8. $\frac{24wz^2}{3w^3z^5}$

   Solution: For this expression, we simplify the coefficients by division and subtract the exponents of like terms (numerator - denominator). So, $\frac{24wz^2}{3w^3z^5} = \frac{24}{3} \cdot w^{1-3}z^{2-5} = 8w^{-2}z^{-3}$. This can also be written as $\frac{8}{w^2z^3}$ if you prefer not to have negative exponents.

9. $c^{\frac{2}{3}}$

   Solution: The expression $c^{\frac{2}{3}}$ can be rewritten in radical notation as $\sqrt[3]{c^2}$. This is because the denominator of the fraction in the exponent represents the root (in this case, the cube root), and the numerator represents the power (in this case, squared).

10. $x^{\frac{1}{6}}$

Solution: Similarly, the expression $x^{\frac{1}{6}}$ can be rewritten in radical notation as $\sqrt[6]{x}$. Here, the denominator of the fraction in the exponent is 6, which represents the sixth root, and the numerator is 1, which represents the first power (or simply x).

17. $\sqrt{16x^{13}}$

Solution: The square root of 16 is 4. For $x^{13}$, we can take out pairs of x's as $x^2$. There are 6 pairs and one x left inside the square root. So, $\sqrt{16x^{13}} = 4x^6\sqrt{x}$.

18. $\sqrt[3]{27a^6b^2}$

Solution: The cube root of 27 is 3. For $a^6$, we can take out pairs of $a^3$ as there are 2 pairs. For $b^2$, there aren't enough b's to take out of the cube root, so it stays inside. So, $\sqrt[3]{27a^6b^2} = 3a^2\sqrt[3]{b^2}$.

19. $\sqrt{(3c)^9}$

Solution: When taking the square root of a term raised to an odd power, we can take out half of the power (rounded down) and leave the rest inside the square root. The square root of $3^9$ is $3^4$ (since $4.5$ rounds down to $4$), and the square root of $c^9$ is $c^4\sqrt{c}$. So, $\sqrt{(3c)^9} = 3^4c^4\sqrt{3c} = 81c^4\sqrt{3c}$.

21. $\sqrt{2} + \sqrt{50} - 4\sqrt{8}$

Solution: We can simplify each term first. $\sqrt{50}$ simplifies to $5\sqrt{2}$ and $\sqrt{8}$ simplifies to $2\sqrt{2}$. Substituting these simplifications back into the original expression gives us $\sqrt{2} + 5\sqrt{2} - 4*2\sqrt{2}$. Combining like terms, we find that the expression simplifies to $0$.

22. $4\left(2\sqrt{9} - 5\right)$

Solution: We can simplify inside the parentheses first. $2\sqrt{9}$ simplifies to $23 = 6$. Substituting this simplification back into the original expression gives us $4\left(6 - 5\right)$, which simplifies to $4(1) = 4$.

24. $(1+2\sqrt{7})(5 + \sqrt{7})$

Solution: We can use the distributive property (also known as the FOIL method) to expand this expression.

First, we multiply the first terms in each binomial: $1 \cdot 5 = 5$.

Next, we multiply the outer terms: $1 \cdot \sqrt{7} = \sqrt{7}$.

Then, we multiply the inner terms: $2\sqrt{7} \cdot 5 = 10\sqrt{7}$.

Finally, we multiply the last terms: $2\sqrt{7} \cdot \sqrt{7} = 2 \cdot 7 = 14$.

Adding these results together, we get $5 + \sqrt{7} + 10\sqrt{7} + 14 = 19 + 11\sqrt{7}$.

27. Find the area of the triangle given the following characteristics: $b = 10x$, $h = 4x^2$, $A = \frac{1}{2}bh$

Solution: We can substitute the given values of $b$ and $h$ into the formula for the area of a triangle, which is $A = \frac{1}{2}bh$.

Substituting $b = 10x$ and $h = 4x^2$ into the formula gives us $A = \frac{1}{2} \cdot 10x \cdot 4x^2 = 20x^3$.

So, the area of the triangle is $20x^3$ square units.

28. $\sqrt{-144}$

Solution: The square root of a negative number is an imaginary number. The square root of -144 is equal to the square root of 144 times the square root of -1. The square root of 144 is 12, and the square root of -1 is denoted as $i$. So, $\sqrt{-144} = 12i$.

29. $i^3$

Solution: The power of $i$ cycles every four powers: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^5 = i$, and so on. So, $i^3 = -i$. Therefore, $9i^3 = 9(-i) = -9i$.

30. $(6i - 8) + (-3i + 2)$

Solution: We can combine like terms to simplify this expression. $6i - 3i = 3i$ and $-8 + 2 = -6$. So, $(6i - 8) + (-3i + 2) = 3i - 6$, or $3(-2 + i)$ in standard form.

31. $4i(i+2)$

Solution: We can distribute $4i$ to both terms inside the parentheses: $4i \cdot i = 4i^2$ and $4i \cdot 2 = 8i$. Remember that $i^2 = -1$, so $4i^2 = 4(-1) = -4$. Therefore, $4i(i+2) = -4 + 8i$, or $4(-1 + 2i)$ in standard form.

34. $\frac{12}{3-i}$

Solution: To simplify a complex fraction, we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $3-i$ is $3+i$. So, we multiply the fraction by $\frac{3+i}{3+i}$ to get $\frac{12(3+i)}{(3-i)(3+i)}$. The denominator simplifies to $3^2-(-i)^2 = 9-(-1) = 10$. So, the fraction simplifies to $\frac{6(3+i)}{5}$, or $3.6+1.2i$ in decimal form.

35. $\sqrt{-2} \cdot \sqrt{-9}$

Solution: The square root of a negative number is an imaginary number. The square root of -2 is $i\sqrt{2}$, and the square root of -9 is $3i$. So, $\sqrt{-2} \cdot \sqrt{-9} = i\sqrt{2} \cdot 3i = -3\sqrt{2}$. The negative sign comes from the fact that $i \cdot i = i^2 = -1$.

36. $f(x) = -\sqrt{x} + 3 + 2$

Solution:

Transformations:

• The negative sign in front of the square root function reflects the graph over the x-axis.
• The "+3" inside the square root function shifts the graph 3 units to the left.
• The "+2" outside the square root function shifts the graph 2 units up.

To graph this function, start with the graph of $y = \sqrt{x}$, then apply the transformations in the order listed above.

37. $f(x) = 2\sqrt{x} - 1 - 4$

Solution:

Transformations:

• The "2" in front of the square root function vertically stretches the graph by a factor of 2.
• The "-1" inside the square root function shifts the graph 1 unit to the right.
• The "-4" outside the square root function shifts the graph 4 units down.

To graph this function, start with the graph of $y = \sqrt{x}$, then apply the transformations in the order listed above.

Note: For both functions, the transformations are applied in the order of reflection, horizontal shift, and then vertical shift.

38. $\sqrt{x} + 3 = 5$

Solution: To solve this equation, we first isolate the square root term by subtracting 3 from both sides of the equation: $\sqrt{x} = 5 - 3 = 2$. Then, we square both sides to eliminate the square root: $x = 2^2 = 4$.

39. $3\sqrt{x} - 2 = 7$

Solution: First, we isolate the square root term by adding 2 to both sides of the equation: $3\sqrt{x} = 7 + 2 = 9$. Then, we divide both sides by 3 to get $\sqrt{x} = \frac{9}{3} = 3$. Finally, we square both sides to eliminate the square root: $x = 3^2 = 9$.

Units 9 and 10: EXPONENTIAL & LOGARITHMIC FUNCTIONS

[example video][https://youtu.be/H46QVUdbBn0] [practice problem][https://www.ixl.com/math/algebra-2/graph-exponential-functions] [caracteristics of function graph][https://youtu.be/woVwBmwhA2Y]

Exponential Form & Logarithmic Form

Identify the characteristics and graph the following functions.

$f(x) = 3^x + 1$

Characteristics:

• Type: Growth
• Domain: All real numbers
• Range: All real numbers greater than or equal to 1

Graph:

To graph the function, we can plot a few points to get an idea of the shape of the graph. Let's choose some x-values and calculate the corresponding y-values:

For $x = -2$: $f(-2) = 3^{-2} + 1 = \frac{1}{3^2} + 1 = \frac{1}{9} + 1 = \frac{10}{9}$

For $x = -1$: $f(-1) = 3^{-1} + 1 = \frac{1}{3^1} + 1 = \frac{1}{3} + 1 = \frac{4}{3}$

For $x = 0$: $f(0) = 3^0 + 1 = 1 + 1 = 2$

For $x = 1$: $f(1) = 3^1 + 1 = 3 + 1 = 4$

For $x = 2$: $f(2) = 3^2 + 1 = 9 + 1 = 10$

Now, let's plot these points on a graph and connect them to get the shape of the exponential function $f(x) = 3^x + 1$. Remember to label the axes and indicate any asymptotes if applicable.

The graph of the function $f(x) = 3^x + 1$ should resemble an increasing curve that approaches the x-axis but never touches or crosses it. The point (0, 2) is the y-intercept of the graph.

Identify the characteristics and graph the following functions.

$f(x) = \log_{\frac{1}{2}}(x)$

Characteristics:

• Vertical Asymptote (VA): $x = 0$
• x-intercept (X-Int): $x = 1$
• Domain: $x > 0$
• Range: All real numbers

Graph:

To graph the function, we can plot some points and draw the graph:

• For $x = \frac{1}{8}$: $f\left(\frac{1}{8}\right) = \log_{\frac{1}{2}}\left(\frac{1}{8}\right) = -3$

• For $x = \frac{1}{4}$: $f\left(\frac{1}{4}\right) = \log_{\frac{1}{2}}\left(\frac{1}{4}\right) = -2$

• For $x = \frac{1}{2}$: $f\left(\frac{1}{2}\right) = \log_{\frac{1}{2}}\left(\frac{1}{2}\right) = -1$

• For $x = 1$: $f(1) = \log_{\frac{1}{2}}(1) = 0$

• For $x = 2$: $f(2) = \log_{\frac{1}{2}}(2) = 1$

• For $x = 4$: $f(4) = \log_{\frac{1}{2}}(4) = 2$

• For $x = 8$: $f(8) = \log_{\frac{1}{2}}(8) = 3$

By plotting these points and considering the vertical asymptote at $x = 0$, we can sketch the graph of the function $f(x) = \log_{\frac{1}{2}}(x)$.

Remember to label the axes and indicate any asymptotes or intercepts on the graph.

Write an exponential function for the graph that passes through the given points. Then determine if the function is growth or decay.

Given Points:

• Point 1: (0, 3)
• Point 2: (2, 192)

To write an exponential function that passes through the given points, we'll use the general form of an exponential function:

$f(x) = ab^x$

where $a$ represents the initial value or the value of the function when $x = 0$, and $b$ represents the base of the exponential function.

1. Finding the value of $a$: Using the point (0, 3), we have:

$3 = ab^0$

Since any number raised to the power of 0 is equal to 1, we can simplify to:

$3 = a \cdot 1$ $a = 3$

Therefore, the value of $a$ is 3.

2. Finding the value of $b$: Using the point (2, 192), we have:

$192 = 3b^2$

Dividing both sides by 3:

$64 = b^2$

Taking the square root of both sides:

$b = \pm 8$

Since exponential functions can't have a negative base, we'll take the positive value:

$b = 8$

Therefore, the exponential function that passes through the given points is:

$f(x) = 3 \cdot 8^x$

To determine whether the function is growth or decay, we observe that the base $b = 8$ is greater than 1. Therefore, the function represents growth.

Write an exponential function for the graph that passes through the given points. Then determine if the function is growth or decay.

Given Points:

• Point 1: (0, 20)
• Point 2: (2, 5)

To write an exponential function that passes through the given points, we'll use the general form of an exponential function:

$f(x) = ab^x$

where $a$ represents the initial value or the value of the function when $x = 0$, and $b$ represents the base of the exponential function.

1. Finding the value of $a$: Using the point (0, 20), we have:

$20 = ab^0$

Since any number raised to the power of 0 is equal to 1, we can simplify to:

$20 = a \cdot 1$ $a = 20$

Therefore, the value of $a$ is 20.

2. Finding the value of $b$: Using the point (2, 5), we have:

$5 = 20b^2$

Dividing both sides by 20:

$\frac{1}{4} = b^2$

Taking the square root of both sides:

$b = \pm \frac{1}{2}$

Since exponential functions can't have a negative base, we'll take the positive value:

$b = \frac{1}{2}$

Therefore, the exponential function that passes through the given points is:

$f(x) = 20 \left(\frac{1}{2}\right)^x$

To determine whether the function is growth or decay, we observe that the base $b = \frac{1}{2}$ is between 0 and 1. Therefore, the function represents decay.

Write an exponential function for the graph that passes through the given points. Then determine if the function is growth or decay.

Given Points:

• Point 1: (0, 1)
• Point 2: (3, 216)

To write an exponential function that passes through the given points, we'll use the general form of an exponential function:

$f(x) = ab^x$

where $a$ represents the initial value or the value of the function when $x = 0$, and $b$ represents the base of the exponential function.

1. Finding the value of $a$: Using the point (0, 1), we have:

$1 = ab^0$

Since any number raised to the power of 0 is equal to 1, we can simplify to:

$1 = a \cdot 1$ $a = 1$

Therefore, the value of $a$ is 1.

2. Finding the value of $b$: Using the point (3, 216), we have:

$216 = 1 \cdot b^3$

Simplifying:

$b^3 = 216$

Taking the cube root of both sides:

$b = 6$

Therefore, the exponential function that passes through the given points is:

$f(x) = 1 \cdot 6^x$

To determine whether the function is growth or decay, we observe that the base $b = 6$ is greater than 1. Therefore, the function represents growth.

The equation $3^{-2x + 10} = 27^{6x - 10}$ can be solved using the method of common base as follows:

First, we need to express both sides of the equation in terms of the same base. We know that $3^3 = 27$, so we can rewrite $27$ as $3^3$. This gives us:

$$3^{-2x + 10} = (3^3)^{6x - 10}$$

Simplifying the right side of the equation gives us:

$$3^{-2x + 10} = 3^{18x - 30}$$

Since the bases are the same, we can equate the exponents:

$$-2x + 10 = 18x - 30$$

Solving this equation for $x$ gives us the real solution:

$$x = 2$$

Here is a plot of the equation, which confirms that $x = 2$ is the solution:

For more details, you can check the Wolfram|Alpha result.

Please note that the equation also has complex solutions, but since we are dealing with real numbers, we only consider the real solution.

The equation $4^{3-2x} = 64^x$ can be solved using the method of common base as follows:

First, we need to express both sides of the equation in terms of the same base. We know that $4^3 = 64$, so we can rewrite $64$ as $4^3$. This gives us:

$$4^{3-2x} = (4^3)^x$$

Simplifying the right side of the equation gives us:

$$4^{3-2x} = 4^{3x}$$

Since the bases are the same, we can equate the exponents:

$$3-2x = 3x$$

Solving this equation for $x$ gives us the real solution:

$$x = \frac{3}{5}$$

Here is a plot of the equation, which confirms that $x = \frac{3}{5}$ is the solution:

For more details, you can check the Wolfram|Alpha result.

Please note that the equation also has complex solutions, but since we are dealing with real numbers, we only consider the real solution.

The equation $b^x = k$ can be rewritten in logarithmic form as follows:

The base $b$ becomes the base of the logarithm, the exponent $x$ is the result, and $k$ is the argument of the logarithm. Therefore, the equation in logarithmic form is:

$$\log_b k = x$$

This is read as "log base $b$ of $k$ equals $x$". It means the same thing as the original equation: "$b$ raised to the power of $x$ equals $k$".

The equation $3^{6x} = 12$ can be rewritten in logarithmic form as follows:

The base $3$ becomes the base of the logarithm, the exponent $6x$ is the result, and $12$ is the argument of the logarithm. Therefore, the equation in logarithmic form is:

$$\log_3 12 = 6x$$

This is read as "log base $3$ of $12$ equals $6x$". It means the same thing as the original equation: "$3$ raised to the power of $6x$ equals $12$".

The equation $\log_8 x = y$ can be rewritten in exponential form as follows:

The base of the logarithm $8$ becomes the base of the exponent, the result $y$ is the exponent, and $x$ is the result of the exponentiation. Therefore, the equation in exponential form is:

$$8^y = x$$

This is read as "8 raised to the power of $y$ equals $x$". It means the same thing as the original equation: "log base $8$ of $x$ equals $y$".

The equation $\log_b t = p$ can be rewritten in exponential form as follows:

The base of the logarithm $b$ becomes the base of the exponent, the result $p$ is the exponent, and $t$ is the result of the exponentiation. Therefore, the equation in exponential form is:

$$b^p = t$$

This is read as "$b$ raised to the power of $p$ equals $t$". It means the same thing as the original equation: "log base $b$ of $t$ equals $p$".

The logarithm $\log_2 16$ is asking the question: "What power do we need to raise $2$ to in order to get $16$?"

The answer to this is $4$, because $2^4 = 16$.

Here is a visual representation of the number $4$:

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For more details, you can check the Wolfram|Alpha result.

The logarithm $\log_{64} \frac{1}{4}$ is asking the question: "What power do we need to raise $64$ to in order to get $\frac{1}{4}$?"

The answer to this is $-\frac{1}{3}$, because $64^{-\frac{1}{3}} = \frac{1}{4}$.

Here is a number line representation of the number $-\frac{1}{3}$:

For more details, you can check the Wolfram|Alpha result.

The logarithm $\log_{\frac{1}{7}} \frac{1}{343}$ is asking the question: "What power do we need to raise $\frac{1}{7}$ to in order to get $\frac{1}{343}$?"

The answer to this is $3$, because $\left(\frac{1}{7}\right)^3 = \frac{1}{343}$.

Here is a visual representation of the number $3$:

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For more details, you can check the Wolfram|Alpha result.

The logarithm $\log_{27} 3$ is asking the question: "What power do we need to raise $27$ to in order to get $3$?"

The answer to this is $\frac{1}{3}$, because $27^{\frac{1}{3}} = 3$.

Here is a number line representation of the number $\frac{1}{3}$:

For more details, you can check the Wolfram|Alpha result.

1. Rational Expressions: Introduction to Rational Expressions by Khan Academy
2. Multiplying and Dividing Rational Expressions: Multiplying and Dividing Rational Expressions by Mathispower4u
3. Adding and Subtracting Rational Expressions: Adding and Subtracting Rational Expressions by Khan Academy
4. Solving Rational Equations: Solving Rational Equations by The Organic Chemistry Tutor
5. Graphing Rational Functions: How to graph a rational function using 6 steps by Brian McLogan

Here are the steps to multiply the rational expressions $\frac{m^2+8m+15}{a+2b}$ and $\frac{7a+14b}{m+3}$:

1. Identify the Numerator and Denominator of Each Fraction: The first fraction has $m^2+8m+15$ as its numerator and $a+2b$ as its denominator. The second fraction has $7a+14b$ as its numerator and $m+3$ as its denominator.

2. Multiply the Numerators: Multiply $(m^2+8m+15)$ and $(7a+14b)$ together. This gives us $(7a+14b)(m^2+8m+15)$ as the numerator of the product.

3. Multiply the Denominators: Multiply $(a+2b)$ and $(m+3)$ together. This gives us $(a+2b)(m+3)$ as the denominator of the product.

4. Write the Product as a Single Fraction: The product of the two fractions is then $\frac{(7a+14b)(m^2+8m+15)}{(a+2b)(m+3)}$.

This is the simplified form of the product of the two given rational expressions. It's important to note that we cannot simplify this expression further without knowing the values of $a$, $b$, and $m$.

Here are the steps to multiply the rational expressions $\frac{6r+3}{r+6}$ and $\frac{r^2+9r+18}{2r+1}$:

1. Identify the Numerator and Denominator of Each Fraction: The first fraction has $6r+3$ as its numerator and $r+6$ as its denominator. The second fraction has $r^2+9r+18$ as its numerator and $2r+1$ as its denominator.

2. Multiply the Numerators: Multiply $(6r+3)$ and $(r^2+9r+18)$ together. This gives us $(6r+3)(r^2+9r+18)$ as the numerator of the product.

3. Multiply the Denominators: Multiply $(r+6)$ and $(2r+1)$ together. This gives us $(r+6)(2r+1)$ as the denominator of the product.

4. Write the Product as a Single Fraction: The product of the two fractions is then $\frac{(6r+3)(r^2+9r+18)}{(r+6)(2r+1)}$.

This is the simplified form of the product of the two given rational expressions. It's important to note that we cannot simplify this expression further without knowing the value of $r$.

Here are the steps to divide the rational expressions $\frac{x}{x+2}$ and $\frac{x^2}{x^2+6x+8}$:

1. Identify the Numerator and Denominator of Each Fraction: The first fraction has $x$ as its numerator and $x+2$ as its denominator. The second fraction has $x^2$ as its numerator and $x^2+6x+8$ as its denominator.

2. Reciprocal of the Second Fraction: When dividing fractions, we multiply by the reciprocal of the second fraction. The reciprocal of $\frac{x^2}{x^2+6x+8}$ is $\frac{x^2+6x+8}{x^2}$.

3. Multiply the Fractions: Multiply $\frac{x}{x+2}$ and $\frac{x^2+6x+8}{x^2}$. This gives us $\frac{x(x^2+6x+8)}{x^2(x+2)}$ as the result.

4. Simplify the Result: The $x^2$ in the numerator and denominator cancel out, leaving us with $\frac{x^2+6x+8}{x(x+2)}$.

This is the simplified form of the quotient of the two given rational expressions. It's important to note that we cannot simplify this expression further without knowing the value of $x$.

Here are the steps to add and subtract the rational expressions $\frac{2}{x}$, $-\frac{1}{4}$, and $\frac{3}{2x}$:

1. Identify Like Terms: In this case, $\frac{2}{x}$ and $\frac{3}{2x}$ are like terms because they have the same variable part, $x$ in the denominator. The term $-\frac{1}{4}$ is a constant and does not have a variable part.

2. Add/Subtract Like Terms: Add the coefficients of the like terms together. The coefficient of $\frac{2}{x}$ is $2$ and the coefficient of $\frac{3}{2x}$ is $\frac{3}{2}$. Adding these together gives $2 + \frac{3}{2} = \frac{7}{2}$, so the sum of these two terms is $\frac{7}{2x}$.

3. Combine the Results: Combine the result from step 2 with the constant term $-\frac{1}{4}$. This gives the final result of $-\frac{1}{4} + \frac{7}{2x}$.

This is the simplified form of the sum of the three given rational expressions. It's important to note that we cannot simplify this expression further without knowing the value of $x$.

The difference of the rational expressions $\frac{4x+1}{x^2-4}$ and $\frac{3}{x-2}$ is:

$$\frac{4x+1}{x^2-4} - \frac{3}{x-2} = \frac{4x+1}{(x-2)(x+2)} - \frac{3}{x-2}$$

This is the simplified form of the difference of the two given rational expressions. Now, let's simplify further:

1. Identify Common Denominator: The common denominator of the two fractions is $(x-2)(x+2)$.

2. Rewrite the Second Fraction: Rewrite the second fraction $\frac{3}{x-2}$ as $\frac{3(x+2)}{(x-2)(x+2)}$ to have the common denominator.

3. Subtract the Fractions: Now that the fractions have the same denominator, subtract the numerators: $\frac{4x+1-3(x+2)}{(x-2)(x+2)}$.

4. Simplify the Numerator: Simplify the numerator to get the final result.

Without knowing the value of $x$, we cannot simplify this expression further.

The difference of the rational expressions $\frac{3-4x}{x^2+3x-10}$ and $\frac{2}{x+5}$ is:

$$\frac{3-4x}{x^2+3x-10} - \frac{2}{x+5} = -\frac{2}{x+5} + \frac{3-4x}{x^2+3x-10}$$

This is the simplified form of the difference of the two given rational expressions. Now, let's simplify further:

1. Identify Common Denominator: The common denominator of the two fractions is $x^2+3x-10$.

2. Rewrite the Second Fraction: Rewrite the second fraction $\frac{2}{x+5}$ as $\frac{2(x-2)}{x^2+3x-10}$ to have the common denominator.

3. Subtract the Fractions: Now that the fractions have the same denominator, subtract the numerators: $-\frac{2(x-2)}{x^2+3x-10} + \frac{3-4x}{x^2+3x-10}$.

4. Simplify the Numerator: Simplify the numerator to get the final result.

Without knowing the value of $x$, we cannot simplify this expression further.

The sum of the rational expressions $\frac{-3x}{x-3}$ and $\frac{4}{2x-6}$ is:

$$\frac{-3x}{x-3} + \frac{4}{2x-6} = -\frac{3x}{x-3} + \frac{4}{2(x-3)}$$

This is the simplified form of the sum of the two given rational expressions. Now, let's simplify further:

1. Identify Common Denominator: The common denominator of the two fractions is $x-3$.

2. Rewrite the Second Fraction: Rewrite the second fraction $\frac{4}{2x-6}$ as $\frac{2 \cdot 4}{2(x-3)}$ to have the common denominator.

3. Add the Fractions: Now that the fractions have the same denominator, add the numerators: $-\frac{3x}{x-3} + \frac{2 \cdot 4}{2(x-3)}$.

4. Simplify the Numerator: Simplify the numerator to get the final result.

Without knowing the value of $x$, we cannot simplify this expression further.

The sum of the rational expressions $\frac{5}{x^2-x-6}$ and $\frac{4}{x^2+4x+4}$ is:

$$\frac{5}{x^2-x-6} + \frac{4}{x^2+4x+4} = \frac{5}{-6 - x + x^2} + \frac{4}{4 + 4x + x^2}$$

This is the simplified form of the sum of the two given rational expressions. Now, let's simplify further:

1. Identify Common Denominator: The common denominator of the two fractions is $x^2-x-6$ and $x^2+4x+4$.

2. Rewrite the Fractions: Rewrite the fractions as $\frac{5}{-6 - x + x^2}$ and $\frac{4}{4 + 4x + x^2}$ to have the common denominator.

3. Add the Fractions: Now that the fractions have the same denominator, add the numerators.

4. Simplify the Numerator: Simplify the numerator to get the final result.

Without knowing the value of $x$, we cannot simplify this expression further.

Quizzes

Throughout the module, there will be quizzes to test your knowledge on the topics we have covered. These quizzes will help you assess your understanding and identify areas where you may need to review.

Active Learning Strategies

During the virtual lecture, we will use active learning strategies to engage with the material and deepen our understanding. These strategies may include group discussions, problem-solving activities, and interactive simulations.

Conclusion

By the end of this module, you should have a strong foundation in Algebra 1 and be able to apply your knowledge to solve real-world problems.

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