The math teacher asked his students to determine a solution for 1.5z2y8. Which solution below is the correct one? (1 point)

Question

Bem

Item

2

W

9

S

a

Fiona believes (4, 7) is a solution

Pater believes (-4, 7) is a solution.

Sam believes (-4, -7) is a solution.

Mary believes (4, -7) is a solution.

Answer 1

To determine whether the points proposed by the students are solutions to the equation \( 1.5z^2y^8 \), we first need to understand what the equation describes. However, since the equation is not set equal to anything, it seems we might be looking for values of \( z \) and \( y \) that make \( 1.5z^2y^8 \) equal to some constant, often \(0\).

From the options given:

1. Fiona believes (4, 7) is a solution:

   • \( z = 4 \) and \( y = 7 \)
   • \( 1.5(4^2)(7^8) > 0 \)

2. Pater believes (-4, 7) is a solution:

   • \( z = -4 \) and \( y = 7 \)
   • \( 1.5((-4)^2)(7^8) > 0 \)

3. Sam believes (-4, -7) is a solution:

   • \( z = -4 \) and \( y = -7 \)
   • \( 1.5((-4)^2)(-7^8) < 0\), since \( (-7)^8 \) is positive but the overall expression will be negative due to the \( -7 \).

4. Mary believes (4, -7) is a solution:

   • \( z = 4 \) and \( y = -7 \)
   • \( 1.5(4^2)(-7^8) < 0\), since \( (-7)^8 \) is positive but the overall expression will be negative due to the \( -7 \).

Assuming we're looking for any positive solution to \( 1.5z^2y^8 \), both Fiona and Pater's answers yield positive results. Given that it doesn't appear there are instructions to specifically identify all possible values but rather looking for solutions that yield a certain output, it's critical to know whether the teacher was looking for positive, negative results, or zero.

If the goal was positivity, both Fiona and Pater are correct. If no restrictions are specified, Fiona (4, 7) or Pater (-4, 7) would suffice as acceptable solutions, with Mary and Sam yielding negative outcomes.

For correctness definitively, if we needed one specific response based on typical interpretations of equations with two variables usually suggesting interest in positive outcomes, I would lean on Fiona believes (4, 7) is a solution as the baseline conventional answer.

Answer 2

Which of the following ordered pairs is part of the solution set for the equation y 5x^2-2

(1 point)

(-1,43)

<

(-1,5)

Ο (-1,3)

(1,5)

Answer 3

To determine which of the ordered pairs is part of the solution set for the equation \( y = 5x^2 - 2 \), we can substitute the x-values from the ordered pairs into the equation and see if the corresponding y-value matches.

1. For the pair (-1, 43): \[ y = 5(-1)^2 - 2 = 5(1) - 2 = 5 - 2 = 3 \] So \((-1, 43)\) is not a solution.

2. For the pair (-1, 5): \[ y = 5(-1)^2 - 2 = 5(1) - 2 = 5 - 2 = 3 \] So \((-1, 5)\) is not a solution.

3. For the pair (-1, 3): \[ y = 5(-1)^2 - 2 = 5(1) - 2 = 5 - 2 = 3 \] So \((-1, 3)\) is a solution.

4. For the pair (1, 5): \[ y = 5(1)^2 - 2 = 5(1) - 2 = 5 - 2 = 3 \] So \((1, 5)\) is not a solution.

The only ordered pair that is part of the solution set for the equation is \((-1, 3)\).

Answer 4

What was the original mass of the Bismuth-212 sample if, after 121 seconds, it now weighs 2.64 grams, given that Bismuth-212 has a half-life of 60.5 seconds? (1 point)

10.56 g

0.66 g

5.28 g

1.32 g

Answer 5

To find the original mass of the Bismuth-212 sample, we use the formula for exponential decay associated with half-lives. The formula is:

\[ m(t) = m_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

Where:

• \( m(t) \) is the remaining mass after time \( t \),
• \( m_0 \) is the original mass,
• \( t \) is the elapsed time (121 seconds in this case),
• \( T_{1/2} \) is the half-life of the substance (60.5 seconds for Bismuth-212).

We know \( m(t) = 2.64 \) grams, \( t = 121 \) seconds, and \( T_{1/2} = 60.5 \) seconds.

First, calculate the number of half-lives that have elapsed in 121 seconds:

\[ n = \frac{t}{T_{1/2}} = \frac{121 \text{ seconds}}{60.5 \text{ seconds}} \approx 2 \]

This means that approximately 2 half-lives have passed.

Next, we can express the remaining mass in terms of the original mass by recognizing that 2 half-lives will reduce the original mass by a factor of \( \left(\frac{1}{2}\right)^2 \):

\[ m(t) = m_0 \left(\frac{1}{2}\right)^{2} = m_0 \left(\frac{1}{4}\right) \]

Now substitute \( m(t) \) into the equation:

\[ 2.64 = m_0 \left(\frac{1}{4}\right) \]

To find \( m_0 \):

\[ m_0 = 2.64 \times 4 = 10.56 \text{ grams} \]

Thus, the original mass of the Bismuth-212 sample was 10.56 grams.