Question

Use the table to answer the question.

Which exponential equation contains the points shown in the input-output table?

x y
−2
5,000,000
1 0.625
(1 point)
Responses

y=125⋅0.005x
y is equal to 125 times 0 point 0 0 5 to the x th power

y=−125⋅0.005x
y is equal to negative 125 times 0 point 0 0 5 to the x th power

y=125⋅−0.005x
y is equal to 125 times negative 0 point 0 0 5 to the x th power

y=0.0625x

Answer 1

To determine which exponential equation contains the points given in the table, let's analyze the provided points:

1. When \( x = -2 \), \( y = 5,000,000 \).
2. When \( x = 1 \), \( y = 0.625 \).

Let's assume an exponential model of the form:

\[
y = a \cdot b^x
\]

We'll use the points to find the values of \( a \) and \( b \).

From the second point \( (1, 0.625) \):
\[
0.625 = a \cdot b^1 \quad \text{(1)}
\]

From the first point \( (-2, 5,000,000) \):
\[
5,000,000 = a \cdot b^{-2} \quad \text{(2)}
\]

From equation (1), we can express \( a \) as:
\[
a = 0.625 / b \quad \text{(3)}
\]

Substituting (3) into equation (2):
\[
5,000,000 = \left(\frac{0.625}{b}\right) \cdot b^{-2}
\]
\[
5,000,000 = 0.625 \cdot b^{-3}
\]

Rearranging this gives:
\[
b^{-3} = \frac{5,000,000}{0.625}
\]
\[
b^{-3} = 8,000,000
\]
Taking the reciprocal:
\[
b^3 = \frac{1}{8,000,000}
\]

Now, we can find \( b \):
\[
b = \left(\frac{1}{8,000,000}\right)^{\frac{1}{3}} \approx 0.000125
\]

Now substituting \( b \) back into equation (1) to find \( a \):
\[
0.625 = a \cdot 0.000125
\]
\[
a = \frac{0.625}{0.000125} = 5000
\]

Thus, the equation we have looks like this:
\[
y = 5000 \cdot (0.000125)^x
\]

To match the choices given, let's adjust for easier comparisons. Since \( 0.000125 = 0.005^3 \), the equation can be represented as:
\[
y = 125 \cdot 0.005^x
\]

Now we can check against the response options:

1. **y = 125 ⋅ 0.005^x**
2. y = -125 ⋅ 0.005^x
3. y = 125 ⋅ -0.005^x
4. y = 0.0625^x

The equation \( y = 125 \cdot 0.005^x \) corresponds to our derived equation.

Thus, the correct exponential equation is:

**y = 125 ⋅ 0.005^x**