Abby and Anita receive water service from two different companies. Anita’s water service costs 30 cents per gallon per month and has a service charge of $25 per month. Abby’s water service costs 10 cents per gallon per month and has a service charge of $35 per month. Let x

Question

represent the number of gallons of water per month and y
represent the monthly cost. Write the equations that make up the system of equations for this scenario.(1 point)
Anita’s water bill can be modeled with the equation y=


Abby’s water bill can be modeled with the equation y=

Answer 1

To write the equations for both Anita's and Abby's water bills, we can break down their costs based on their pricing structures.

1. Anita's Water Bill:

   • The variable \( x \) represents the number of gallons of water used.
   • The cost per gallon is 30 cents, which can be represented as \( 0.30x \).
   • There is also a fixed service charge of $25 per month.
   • Therefore, Anita's total monthly cost \( y \) can be modeled by the equation:

   \[ y = 0.30x + 25 \]

2. Abby's Water Bill:

   • The variable \( x \) represents the number of gallons of water used.
   • The cost per gallon is 10 cents, which can be represented as \( 0.10x \).
   • There is also a fixed service charge of $35 per month.
   • Therefore, Abby's total monthly cost \( y \) can be modeled by the equation:

   \[ y = 0.10x + 35 \]

Hence, the system of equations representing the monthly water costs for Anita and Abby is:

• Anita: \( y = 0.30x + 25 \)
• Abby: \( y = 0.10x + 35 \)

Answer 2

Solve the following quadratic equation. Round to the nearest hundredth if necessary: (x+27)^2/−6=−3
.

Enter the smaller of the 2 values first.

(1 point)
{
}

Answer 3

To solve the equation \(\frac{(x + 27)^2}{-6} = -3\), we first eliminate the fraction by multiplying both sides by \(-6\):

\[ (x + 27)^2 = 18 \]

Next, we take the square root of both sides. Remember that taking the square root will yield both positive and negative solutions:

\[ x + 27 = \pm \sqrt{18} \]

We can simplify \(\sqrt{18}\):

\[ \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \]

So our equation becomes:

\[ x + 27 = 3\sqrt{2} \quad \text{or} \quad x + 27 = -3\sqrt{2} \]

Now we solve for \(x\) in each case:

1. For \(x + 27 = 3\sqrt{2}\):

\[ x = 3\sqrt{2} - 27 \]

2. For \(x + 27 = -3\sqrt{2}\):

\[ x = -3\sqrt{2} - 27 \]

Next, we calculate \(3\sqrt{2}\):

\[ \sqrt{2} \approx 1.414 \] \[ 3\sqrt{2} \approx 3 \times 1.414 \approx 4.242 \]

Now, substituting back into the equations:

1. For the first solution:

\[ x \approx 4.242 - 27 \approx -22.758 \]

2. For the second solution:

\[ x \approx -4.242 - 27 \approx -31.242 \]

We round the results to the nearest hundredth:

1. \(x \approx -22.76\)
2. \(x \approx -31.24\)

Thus, the solutions are:

\[ { -31.24, -22.76 } \]

Entering the smaller value first, we write:

\[ { -31.24, -22.76 } \]

Answer 4

Joana solved the equation 4x2−4x−3=0
. She followed the steps to solve a quadratic equation by grouping. However, in Step 3 she noticed that the two groupings did not share a common binomial, as shown in her work below. She concluded that the quadratic equation has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.

Step 1: 4x2+2x+(−6x)−3=0

Step 2:(4x2+2x)+(−6x−3)=0

Step 3: x(4x+2x)+(−3(2x+1))=0

(1 point)
Responses

No, because she did not factor out the GCF from the first grouping.
No, because she did not factor out the GCF from the first grouping.

No, because she can still continue the steps even if the two groupings have no shared
binomial.
No, because she can still continue the steps even if the two groupings have no shared binomial.

Yes, because there are no factors of −12 that add up to −4.
Yes, because there are no factors of −12 that add up to −4.

Yes, because the two groupings do not share a common binomial.
Yes, because the two groupings do not share a common binomial.

Answer 5

To assess Joana's conclusion regarding the quadratic equation \(4x^2 - 4x - 3 = 0\), let's analyze her steps to see if she is correct.

1. Step 1: Splitting the middle term
   Joana rewrites the equation as \(4x^2 + 2x - 6x - 3 = 0\).
   This is a valid step as she correctly splits \(-4x\) into \(2x\) and \(-6x\).

2. Step 2: Grouping
   She then groups the terms: \((4x^2 + 2x) + (-6x - 3) = 0\).

3. Step 3: Factoring
   Here, Joana attempts to factor each group:

   • From the first group \(4x^2 + 2x\), she gets \(2x(2x + 1)\).
   • From the second group \(-6x - 3\), she factors out \(-3\) but writes \( -3(2x + 1) \).

However, what Joana ended up with in this grouping did not present a common binomial factor.

Key Point:
Just because there isn't a common binomial factor at this step doesn't mean the quadratic equation has no solutions. It indicates an incorrect factoring approach but does not necessarily conclude the equation has no solutions.

Conclusion

The correct response would be:

• No, because she can still continue the steps even if the two groupings have no shared binomial.

This is indicative that the solving process isn't complete. She could utilize the quadratic formula or complete the square to find the solutions. The equation can indeed have solutions, even if her grouping was not successful in isolating a common factor at that stage.

Answer 6

Gwen was asked to solve the equation 3x2−20=7
. What is/are the correct solution(s)?(1 point)
Responses

x=−3
x is equal to negative 3

x=−3, x=3
x=−3, x=3

x=9, x=−9
x=9, x=−9

x=3
x is equal to 3

Answer 7

To solve the equation \(3x^2 - 20 = 7\), we will first isolate the quadratic term.

1. Start by moving 7 to the left side of the equation: \[ 3x^2 - 20 - 7 = 0 \] This simplifies to: \[ 3x^2 - 27 = 0 \]

2. Next, add 27 to both sides: \[ 3x^2 = 27 \]

3. Now, divide both sides by 3: \[ x^2 = 9 \]

4. To find \(x\), take the square root of both sides: \[ x = \pm 3 \]

This gives us two solutions:

• \(x = 3\)
• \(x = -3\)

Thus, the correct solutions are:

Response:
\(x = -3, x = 3\)
\(x = -3, x = 3\)

Answer 8

Which of the following is a solution to the quadratic equation 4x2+2x=10
? (1 point)
Responses

−2+41√/4
the fraction with numerator negative 2 plus square root of 41 and denominator 4

−1−41√/4
the fraction with numerator negative 1 minus square root of 41 and denominator 4

−2+92√/8
the fraction with numerator negative 2 plus 9 square root of 2 and denominator 8

−1−92√/8

Answer 9

To determine which of the given answers is a solution to the quadratic equation \(4x^2 + 2x = 10\), we first want to rewrite it in standard form:

1. Rearranging the equation: \[ 4x^2 + 2x - 10 = 0 \]

2. Applying the Quadratic Formula: The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation \(4x^2 + 2x - 10 = 0\), we have:

   • \(a = 4\)
   • \(b = 2\)
   • \(c = -10\)

   Now we can calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot (-10) = 4 + 160 = 164 \]

3. Finding the square root of the discriminant: \[ \sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41} \]

4. Substituting back into the quadratic formula: \[ x = \frac{-2 \pm 2\sqrt{41}}{2 \cdot 4} = \frac{-2 \pm 2\sqrt{41}}{8} \] Simplifying it gives: \[ x = \frac{-1 \pm \sqrt{41}}{4} \]

Thus, the solutions are: \[ x = \frac{-1 + \sqrt{41}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{41}}{4} \]

Now let's review the provided responses:

1. \( \frac{-2 + 41\sqrt{}}{4} \) - This doesn't match.
2. \( \frac{-1 - \sqrt{41}}{4} \) - This matches our solution!
3. \( \frac{-2 + 9\sqrt{2}}{8} \) - This doesn't match.
4. \( \frac{-1 - 9\sqrt{2}}{8} \) - This doesn't match.

The correct response is: \[ \frac{-1 - \sqrt{41}}{4} \] Thus, Response: \(−1−\frac{\sqrt{41}}{4}\) (the fraction with numerator negative 1 minus square root of 41 and denominator 4) is the solution to the equation.

Answer 10

Which of the following lines is perpendicular to the x-axis?(1 point)
Responses

y=1
y is equal to 1

y=x
y is equal to x

x=1
x is equal to 1

x=−y

Answer 11

To determine which lines are perpendicular to the x-axis, it's essential to understand that a line perpendicular to the x-axis must be a vertical line.

1. Understanding Lines:

   • A vertical line has an equation of the form \(x = k\), where \(k\) is a constant.
   • A horizontal line has an equation of the form \(y = k\).
   • The line \(y = x\) has a slope of 1 and is not vertical or horizontal.
   • The equation \(x = -y\) is also not vertical since it can be rearranged to \(y = -x\), which is a diagonal line with a slope of -1.

2. Analyzing the Options:

   • \(y = 1\): This is a horizontal line.
   • \(y = x\): This is a diagonal line with a slope of 1.
   • \(x = 1\): This is a vertical line where every point on the line has an x-coordinate of 1.
   • \(x = -y\): This can be rewritten as \(y = -x\), which is also a diagonal line (slope of -1).

From the options given, the only line that is vertical and thus perpendicular to the x-axis is:

Response: \(x = 1\) (x is equal to 1).