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Question

Which value of x
is a solution to this equation?

8x2−2x−3=0


(1 point)
Responses

8
8

-0.5
-0.5

-3
-3

4.5
4.5
What is the positive solution to x2−4x−21=0
?(1 point)
Question
The graph shows the height in feet of an object above the ground t
seconds after it was launched from the ground.

Which function is best represented by the graph of this situation?

(1 point)
Responses

h(t)=−16t2−64t
h of t is equal to negative 16 t squared minus 64 t

h(t)=−16t2+128t−256
h of t is equal to negative 16 t squared plus 128 t minus 256

h(t)=−16t2−128t−256
h of t is equal to negative 16 t squared minus 128 t minus 256

h(t)=−16t2+64t
A person dives into a pool from its edge to swim to the other side. The table shows the depth in feet of the person from the surface of the water after x
seconds. The data can be modeled by a quadratic function.

Which function best models the data?

(1 point)
Responses

d(x)=0.05x2+0.74x
d times x is equal to 0 point 0 5 x squared plus 0 point 7 4 x

d(x)=0.05x2+0.74x+9.17
d times x is equal to 0 point 0 5 x squared plus 0 point 7 4 x plus 9 point 1 7

d(x)=0.26x2−3.11x
d times x is equal to 0 point 2 6 x squared minus 3 point 1 1 x

d(x)=0.26x2−3.11x+1

1 answer

Let's break down each question one by one:

  1. Finding the value of x that is a solution to the equation \(8x^2−2x−3=0\):

    To solve \(8x^2 - 2x - 3 = 0\), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 8\), \(b = -2\), and \(c = -3\).

    First, we calculate the discriminant: \[ b^2 - 4ac = (-2)^2 - 4(8)(-3) = 4 + 96 = 100 \]

    Now we substitute into the formula: \[ x = \frac{-(-2) \pm \sqrt{100}}{2(8)} = \frac{2 \pm 10}{16} \]

    This gives us two potential solutions: \[ x = \frac{12}{16} = \frac{3}{4} \quad \text{and} \quad x = \frac{-8}{16} = -\frac{1}{2} \]

    From your responses, none of these are listed. However, if you are looking for negative values, the solution \( -0.5 \) (which is equivalent to \( -\frac{1}{2} \)) is closest.

  2. Finding the positive solution to \(x^2 - 4x - 21 = 0\):

    Again, we can use the quadratic formula with \(a = 1\), \(b = -4\), \(c = -21\). \[ b^2 - 4ac = (-4)^2 - 4(1)(-21) = 16 + 84 = 100 \]

    Now for the values of x: \[ x = \frac{-(-4) \pm \sqrt{100}}{2(1)} = \frac{4 \pm 10}{2} \]

    This results in: \[ x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{-6}{2} = -3 \]

    The positive solution here is \(7\).

  3. Which function is best represented by the graph of the situation for height above ground over time?

    Typically, in free-fall motion, the equation is quadratic and often takes the form \( h(t) = -16t^2 + vt + h_0 \), where \(v\) is the initial velocity and \(h_0\) is the initial height.

    Here, the only function that follows this structure is:

    \[ h(t) = -16t^2 + 128t - 256 \]

  4. Which function best models the depth of a person diving into a pool after \(x\) seconds?

    The function is quadratic, and you have provided several options. The best model would typically involve terms that reflect the shape of a parabolic trajectory, which can vary depending on starting depth and motion.

    The equation that includes the constant term to represent that initial depth would most likely be:

    \[ d(x) = 0.05x^2 + 0.74x + 9.17 \]

So, to summarize:

  1. Solution to \(8x^2 - 2x - 3 = 0\) is \( -0.5 \).
  2. Positive solution to \(x^2 - 4x - 21 = 0\) is \(7\).
  3. The function for the height above ground is \(h(t) = -16t^2 + 128t - 256\).
  4. The model for the diving depth is \(d(x) = 0.05x^2 + 0.74x + 9.17\).