1. which of the following equations has the same solution as $5x + 8 = x - 9$?
a) $4x = -1$ b) $4x = 17$ c) $6x = -17$
d) $6x = 17$ e) $4x = -17$

2. simplify. $(-2x^4)^3(-2x^2)^2$
a) $8x^{11}$ b) $8x^{16}$ c) $-32x^{16}$
d) $32x^{16}$ e) $-32x^{11}$

3. if $f(x) = \sqrt{7 - x}$, then which of the following sets is the domain of this function?
a) $x \leq 7$ b) $x \
eq 7$ c) $x \geq 0$
d) $x \
eq 0$ e) $x \geq 7$

4. one of the factors of $3x^2 + 8x - 35$ is
a) $3x - 7$ b) $3x + 7$ c) $x - 35$
d) $3x + 5$ e) $x - 5$

5. one solution of $3x^2 + 7x - 6 = 0$ is
a) $\frac{-2}{3}$ b) $\frac{3}{2}$ c) $3$
d) $-6$ e) $\frac{2}{3}$

6. solve $\frac{1}{x - 1} - \frac{2}{7} = 3$ for $x$.
a) $\frac{26}{23}$ b) $\frac{-12}{23}$ c) $\frac{23}{26}$
d) $\frac{30}{23}$ e) $\frac{23}{30}$

Question

1. which of the following equations has the same solution as $5x + 8 = x - 9$?
  a) $4x = -1$  b) $4x = 17$  c) $6x = -17$
  d) $6x = 17$  e) $4x = -17$

2. simplify. $(-2x^4)^3(-2x^2)^2$
  a) $8x^{11}$  b) $8x^{16}$  c) $-32x^{16}$
  d) $32x^{16}$  e) $-32x^{11}$

3. if $f(x) = \sqrt{7 - x}$, then which of the following sets is the domain of this function?
  a) $x \leq 7$  b) $x \
eq 7$  c) $x \geq 0$
  d) $x \
eq 0$  e) $x \geq 7$

4. one of the factors of $3x^2 + 8x - 35$ is
  a) $3x - 7$  b) $3x + 7$  c) $x - 35$
  d) $3x + 5$  e) $x - 5$

5. one solution of $3x^2 + 7x - 6 = 0$ is
  a) $\frac{-2}{3}$  b) $\frac{3}{2}$  c) $3$
  d) $-6$  e) $\frac{2}{3}$

6. solve $\frac{1}{x - 1} - \frac{2}{7} = 3$ for $x$.
  a) $\frac{26}{23}$  b) $\frac{-12}{23}$  c) $\frac{23}{26}$
  d) $\frac{30}{23}$  e) $\frac{23}{30}$

Accepted Answer

### Question 1
# Explanation:
## Step1: Solve the given equation
Start with $ 5x + 8 = x - 9 $. Subtract $ x $ from both sides: $ 5x - x + 8 = -9 $, which simplifies to $ 4x + 8 = -9 $. Then subtract 8 from both sides: $ 4x = -9 - 8 $, so $ 4x = -17 $.
# Answer:
E) $ 4x = -17 $

### Question 2
# Explanation:
## Step1: Apply power of a product rule
First, simplify $ (-2x^4)^3 $. Using $ (ab)^n = a^n b^n $ and $ (a^m)^n = a^{mn} $, we get $ (-2)^3 (x^4)^3 = -8x^{12} $.
## Step2: Simplify the second term
Simplify $ (-2x^2)^2 $: $ (-2)^2 (x^2)^2 = 4x^4 $.
## Step3: Multiply the two simplified terms
Multiply $ -8x^{12} $ and $ 4x^4 $: $ (-8 \times 4)x^{12 + 4} = -32x^{16} $.
# Answer:
C) $ -32x^{16} $

### Question 3
# Explanation:
## Step1: Recall the domain of square root function
For a square root function $ \sqrt{f(x)} $, the expression inside the square root (the radicand) must be non - negative, i.e., $ f(x)\geq0 $.
## Step2: Apply to the given function
For $ f(x)=\sqrt{7 - x} $, we need $ 7 - x\geq0 $. Solving for $ x $, we subtract 7 from both sides: $ -x\geq - 7 $, and then multiply both sides by - 1 (remember to reverse the inequality sign) to get $ x\leq7 $.
# Answer:
A) $ x\leq7 $

### Question 4
# Explanation:
## Step1: Factor the quadratic expression
We use the AC method to factor $ 3x^2+8x - 35 $. For a quadratic $ ax^2+bx + c $ (here $ a = 3 $, $ b = 8 $, $ c=-35 $), we need to find two numbers that multiply to $ a\times c=3\times(-35)=-105 $ and add up to $ b = 8 $. The numbers are 15 and - 7.
Rewrite the middle term: $ 3x^2+15x-7x - 35 $.
## Step2: Group and factor
Group the first two and last two terms: $ (3x^2 + 15x)-(7x + 35)=3x(x + 5)-7(x + 5)=(3x - 7)(x + 5) $. So one of the factors is $ 3x - 7 $.
# Answer:
A) $ 3x - 7 $

### Question 5
# Explanation:
## Step1: Factor the quadratic equation
We factor $ 3x^2+7x - 6 = 0 $ using the AC method. $ a = 3 $, $ b = 7 $, $ c=-6 $, $ a\times c=3\times(-6)=-18 $. We need two numbers that multiply to - 18 and add up to 7. The numbers are 9 and - 2.
Rewrite the middle term: $ 3x^2+9x-2x - 6 = 0 $.
## Step2: Group and factor
Group the terms: $ (3x^2 + 9x)-(2x + 6)=3x(x + 3)-2(x + 3)=(3x - 2)(x + 3)=0 $.
## Step3: Solve for x
Set each factor equal to zero: $ 3x - 2 = 0$ gives $ x=\frac{2}{3} $, and $ x + 3=0$ gives $ x=-3 $. So one solution is $ \frac{2}{3} $.
# Answer:
E) $ \frac{2}{3} $

### Question 6
# Explanation:
## Step1: Eliminate the fractions
Multiply each term in the equation $ \frac{1}{x - 1}-\frac{2}{7}=3 $ by $ 7(x - 1) $ (the least common denominator) to get rid of the fractions.
$ 7(x - 1)\times\frac{1}{x - 1}-7(x - 1)\times\frac{2}{7}=3\times7(x - 1) $.
## Step2: Simplify each term
Simplify each term: $ 7-2(x - 1)=21(x - 1) $.
## Step3: Expand and solve for x
Expand the left - hand side: $ 7-2x + 2=21x-21 $. Combine like terms: $ 9-2x=21x-21 $. Add $ 2x $ to both sides: $ 9 = 23x-21 $. Add 21 to both sides: $ 30 = 23x $. Then $ x=\frac{30}{23} $.
# Answer:
D) $ \frac{30}{23} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Solve the given equation

Step1: Apply power of a product rule

Step2: Simplify the second term

Step3: Multiply the two simplified terms

Step1: Recall the domain of square root function

Step2: Apply to the given function

Answer:

Question 2