Question
Which exponential equation contains the points shown in the input-output table?
x y
−2
0.32
1 0.625
(1 point)
Responses
y=0.5⋅(1.25)x
y is equal to 0 point 5 times 1 point 2 5 to the x th power
y=(0.625)x
y is equal to 0 point 6 2 5 to the x th power
y=−0.5⋅(1.25)x
y is equal to negative 0 point 5 times 1 point 2 5 to the x th power
y=0.05⋅−(1.25)x
Answers
GPT-4o mini
To determine which exponential equation fits the provided points from the input-output table, we can analyze the two given points:
1. For \( x = -2 \), \( y = 0.32 \)
2. For \( x = 1 \), \( y = 0.625 \)
We are looking for an equation of the form \( y = a \cdot b^x \).
### Substituting the first point:
Using the point \((-2, 0.32)\):
\[
0.32 = a \cdot b^{-2}
\]
This can be rearranged to:
\[
0.32 = \frac{a}{b^2} \quad (1)
\]
### Substituting the second point:
Using the point \( (1, 0.625) \):
\[
0.625 = a \cdot b^1
\]
This gives us:
\[
0.625 = a \cdot b \quad (2)
\]
### Solving the equations:
From equation (2), we can express \( a \):
\[
a = \frac{0.625}{b} \quad (3)
\]
Now, substitute equation (3) into equation (1):
\[
0.32 = \frac{0.625}{b} \cdot b^{-2}
\]
This simplifies to:
\[
0.32 = \frac{0.625}{b^3}
\]
Multiplying both sides by \( b^3 \):
\[
0.32b^3 = 0.625
\]
Solving for \( b^3 \):
\[
b^3 = \frac{0.625}{0.32} \approx 1.953125
\]
Now, take the cube root of both sides to find \( b \):
\[
b \approx \sqrt[3]{1.953125} \approx 1.25
\]
### Finding \( a \):
Now substitute \( b \approx 1.25 \) back into equation (3) to find \( a \):
\[
a = \frac{0.625}{1.25} = 0.5
\]
### Final Equation:
Now we have:
\[
a = 0.5 \quad \text{and} \quad b = 1.25
\]
Thus, the exponential equation that contains the points is:
\[
y = 0.5 \cdot (1.25)^x
\]
So the correct response is:
**y = 0.5⋅(1.25)x**
1. For \( x = -2 \), \( y = 0.32 \)
2. For \( x = 1 \), \( y = 0.625 \)
We are looking for an equation of the form \( y = a \cdot b^x \).
### Substituting the first point:
Using the point \((-2, 0.32)\):
\[
0.32 = a \cdot b^{-2}
\]
This can be rearranged to:
\[
0.32 = \frac{a}{b^2} \quad (1)
\]
### Substituting the second point:
Using the point \( (1, 0.625) \):
\[
0.625 = a \cdot b^1
\]
This gives us:
\[
0.625 = a \cdot b \quad (2)
\]
### Solving the equations:
From equation (2), we can express \( a \):
\[
a = \frac{0.625}{b} \quad (3)
\]
Now, substitute equation (3) into equation (1):
\[
0.32 = \frac{0.625}{b} \cdot b^{-2}
\]
This simplifies to:
\[
0.32 = \frac{0.625}{b^3}
\]
Multiplying both sides by \( b^3 \):
\[
0.32b^3 = 0.625
\]
Solving for \( b^3 \):
\[
b^3 = \frac{0.625}{0.32} \approx 1.953125
\]
Now, take the cube root of both sides to find \( b \):
\[
b \approx \sqrt[3]{1.953125} \approx 1.25
\]
### Finding \( a \):
Now substitute \( b \approx 1.25 \) back into equation (3) to find \( a \):
\[
a = \frac{0.625}{1.25} = 0.5
\]
### Final Equation:
Now we have:
\[
a = 0.5 \quad \text{and} \quad b = 1.25
\]
Thus, the exponential equation that contains the points is:
\[
y = 0.5 \cdot (1.25)^x
\]
So the correct response is:
**y = 0.5⋅(1.25)x**