Which equation matches the solution set of {5}

Question

?(1 point)
Responses

3x5+4=7
Start Fraction 3 x over 5 End Fraction plus 4 equals 7

x+x−9=1+x
x plus x minus 9 equals 1 plus x

x2−10=15
x squared minus 10 equals 15

2x−20=30

Answer 1

To find the equation that matches the solution set of {5}, we need to evaluate each equation to see if x = 5 is a solution.

1. Equation: \( 3x + 4 = 7 \)

   • Substitute \( x = 5 \): \[ 3(5) + 4 = 15 + 4 = 19 \quad (\text{not } 7) \]
   • This does not match.

2. Equation: \( x + x - 9 = 1 + x \)

   • Simplifying the left side: \[ 2x - 9 = 1 + x \]
   • Substitute \( x = 5 \): \[ 2(5) - 9 = 10 - 9 = 1, \quad 1 + 5 = 6 \quad (\text{not equal}) \]
   • This does not match.

3. Equation: \( x^2 - 10 = 15 \)

   • Rearranging it gives: \[ x^2 = 25 \]
   • Taking the square root: \[ x = 5 \quad \text{or} \quad x = -5 \]
   • This matches one of the solutions, but not just {5}.

4. Equation: \( 2x - 20 = 30 \)

   • Solve for \( x \): \[ 2x = 30 + 20 \] \[ 2x = 50 \] \[ x = 25 \quad (\text{not } 5) \]
   • This does not match.

Among the equations presented, none matches the solution set of exactly {5}. However, \( x^2 - 10 = 15 \) does include 5 as one of its solutions. If the context allows for solutions, this may be the closest match.

If strictly looking for only {5}, then none of the options strictly meet that criteria.

Answer 2

What is the value of p in the following equation? 20/p−3=5
(1 point)
Responses

5
5

7
7

3
3

1

Answer 3

To find the value of \( p \) in the equation

\[ \frac{20}{p} - 3 = 5, \]

let's isolate \( \frac{20}{p} \) first by adding 3 to both sides:

\[ \frac{20}{p} = 5 + 3. \] \[ \frac{20}{p} = 8. \]

Next, by cross-multiplying, we can eliminate the fraction:

\[ 20 = 8p. \]

Now, solve for \( p \) by dividing both sides by 8:

\[ p = \frac{20}{8} = \frac{5}{2} = 2.5. \]

Since 2.5 is not an option among those listed (5, 7, 3, 1), let me check the calculations for you.

The original equation is:

\[ \frac{20}{p} - 3 = 5. \]

Adding 3 to both sides gives:

\[ \frac{20}{p} = 8. \]

Thus:

\[ 20 = 8p \text{ leads to } p = \frac{20}{8} = 2.5. \]

Again, it seems there is some misunderstanding concerning the responses listed. The correct value of \( p \) based on the solution of the equation is 2.5, which is not one of your provided options. Please double-check the options or the equation itself.

Answer 4

which of the options

Answer 5

Based on the solution we found for the equation \(\frac{20}{p} - 3 = 5\), we calculated that \(p = 2.5\). Since 2.5 is not one of the options provided (5, 7, 3, 1), we need to determine which option is the closest, or if perhaps there is an alternative approach.

Let's evaluate the options to see if substituting any of them back into the original equation yields a correct result:

1. Option 5: \[ \frac{20}{5} - 3 = 4 - 3 = 1 \quad (\text{not 5}) \]

2. Option 7: \[ \frac{20}{7} - 3 \approx 2.86 - 3 = -0.14 \quad (\text{not 5}) \]

3. Option 3: \[ \frac{20}{3} - 3 \approx 6.67 - 3 = 3.67 \quad (\text{not 5}) \]

4. Option 1: \[ \frac{20}{1} - 3 = 20 - 3 = 17 \quad (\text{not 5}) \]

None of the options work out correctly in the original equation. Since you asked for which of the provided options aligns best, we can determine that none do.

If forced to choose, based on the original calculation \(p = 2.5\), we might suggest option 3 (3) as it is the closest integer, but again, please note that it is not an exact solution to the equation.

If you have any further context or different options, please let me know!

Answer 6

Get ready for an amusement park day of fun! You and your family want to spend the day at a theme park. You have pooled your money together and have $685. The tickets to the park are $125 each and there is an online processing fee of $5.75 per ticket. You also need to pay $20 for parking. If you have a family of five, do you have enough money to purchase these tickets? Find an inequality to determine how many tickets can be purchased. Then find how many tickets you are able to purchase based on your inequality.(1 point)
Responses

125t≤685
, and yes, you can purchase the tickets.
125 t less than or equal to 685 , and yes, you can purchase the tickets.

125t+5.75t+20≤685
, and yes, you can purchase the tickets.
125 t plus 5.75 t plus 20 less than or equal to 685 , and yes, you can purchase the tickets.

125t+5.75t≤685
, and yes, you can purchase the tickets.
125 t plus 5.75 t less than or equal to 685 , and yes, you can purchase the tickets.

125t+5.75t+20≤685
, and no, you cannot purchase the tick

Answer 7

To determine if you can purchase tickets for your family at the amusement park with a total budget of $685, we need to formulate an inequality that takes into account the cost of the tickets, the processing fee, and the parking fee.

Costs Breakdown:

1. Ticket Cost per Person: $125
2. Processing Fee per Ticket: $5.75
3. Parking Fee: $20

Total Cost per Ticket:

The total cost for each ticket including the processing fee is: \[ 125 + 5.75 = 130.75 \text{ per ticket} \]

Total Equation:

If you denote the number of tickets as \( t \), the total expense will be: \[ Total\ Cost = (125 + 5.75)t + 20 \]

Where \( (125 + 5.75)t \) is the cost of \( t \) tickets and the constant $20 is for parking.

Setting Up the Inequality:

We want to ensure that the total cost does not exceed the amount you have, which is $685: \[ (125 + 5.75)t + 20 \leq 685 \]

Simplifying the Inequality:

This simplifies to: \[ 130.75t + 20 \leq 685 \]

Subtracting $20 from both sides gives: \[ 130.75t \leq 665 \]

Now, solving for \( t \):

\[ t \leq \frac{665}{130.75} \approx 5.08 \]

Since \( t \) must be a whole number, the maximum number of tickets you can purchase is 5.

Conclusion:

Given that the maximum \( t \) is 5, you can purchase 5 tickets.

Final Response:

So, the correct inequality is: \[ 125t + 5.75t + 20 \leq 685, \] and yes, you can purchase the tickets for your family of 5, as you can buy exactly 5 tickets.

Thus, the correct option is: 125t + 5.75t + 20 ≤ 685, and yes, you can purchase the tickets.

Answer 8

Solve x^2=25/64
by inspection. There are two real solutions. Enter the lesser number first. Leave the answers in simplest fraction form. (1 point)
{
,
}

Answer 9

To solve the equation

\[ x^2 = \frac{25}{64} \]

we can take the square root of both sides. When taking the square root, remember that there will be both positive and negative solutions.

Taking the square root gives:

\[ x = \pm \sqrt{\frac{25}{64}}. \]

Now, simplifying the square root:

\[ x = \pm \frac{\sqrt{25}}{\sqrt{64}} = \pm \frac{5}{8}. \]

Thus, the two solutions are:

\[ x = \frac{5}{8} \quad \text{and} \quad x = -\frac{5}{8}. \]

Since we need to list the lesser number first, the final answer is:

\[ \left{-\frac{5}{8}, \frac{5}{8}\right}. \]

Answer 10

Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+27)2/−6=−3
.(1 point)
{
}

Answer 11

To solve the quadratic equation

\[ \frac{(x + 27)^2}{-6} = -3, \]

we can first eliminate the fraction by multiplying both sides of the equation by \(-6\) (noting that this will reverse the equality):

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

Next, we take the square root of both sides:

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

Since \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\), we can rewrite the equation as:

\[ x + 27 = \pm 3\sqrt{2}. \]

Now, we isolate \(x\) by subtracting \(27\) from both sides:

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

Next, we compute the approximate values of the solutions.

First, we need to calculate \(3\sqrt{2}\): \[ \sqrt{2} \approx 1.414, \] thus, \[ 3\sqrt{2} \approx 3 \cdot 1.414 \approx 4.242. \]

Now we can find the two solutions:

1. For the positive root: \[ x_1 = -27 + 3\sqrt{2} \approx -27 + 4.242 \approx -22.758 \quad (\text{rounded to } -22.76) \]

2. For the negative root: \[ x_2 = -27 - 3\sqrt{2} \approx -27 - 4.242 \approx -31.242 \quad (\text{rounded to } -31.24) \]

Thus, rounding to the nearest hundredth, the solutions can be expressed as:

\[ { -31.24, -22.76 }. \]