Asked by bloozoo
Use the table to answer the question.
This table shows the relationship between x and y defined by the given equation along with the corresponding points (x,f(x)) , where f(x) is the inverse of function x in terms of y.
y 3y−1=x
(x,f(x))
−2
−0.89
(−0.89,−2)
−1
−0.67
(−0.67,−1)
0 −1
(−1,B)
1 2 (2,1)
2 A
(A,2)
3 26 (26,3)
In order to graph a logarithmic function, a table has been created. The function has been rewritten in exponential form. What are the values for A
and B
?
(1 point)
A=
and B=
This table shows the relationship between x and y defined by the given equation along with the corresponding points (x,f(x)) , where f(x) is the inverse of function x in terms of y.
y 3y−1=x
(x,f(x))
−2
−0.89
(−0.89,−2)
−1
−0.67
(−0.67,−1)
0 −1
(−1,B)
1 2 (2,1)
2 A
(A,2)
3 26 (26,3)
In order to graph a logarithmic function, a table has been created. The function has been rewritten in exponential form. What are the values for A
and B
?
(1 point)
A=
and B=
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve for \( A \) and \( B \) in the given table, we start with the equation that defines the relationship between \( x \) and \( y \):
\[
3y - 1 = x
\]
We can rearrange this equation to express \( y \) in terms of \( x \):
\[
3y = x + 1 \implies y = \frac{x + 1}{3}
\]
Now we can use this equation to find \( B \) and \( A \).
**Finding \( B \):**
Using the point \( (−1, B) \):
\[
B = \frac{−1 + 1}{3} = \frac{0}{3} = 0
\]
So, we have:
\[
B = 0
\]
**Finding \( A \):**
Using the point \( (A, 2) \):
\[
2 = \frac{A + 1}{3}
\]
Now, we solve for \( A \):
Multiply both sides by \( 3 \):
\[
6 = A + 1
\]
Subtract \( 1 \) from both sides:
\[
A = 6 - 1 = 5
\]
So, we have:
\[
A = 5
\]
In summary, the values are:
\[
A = 5 \quad \text{and} \quad B = 0
\]
\[
3y - 1 = x
\]
We can rearrange this equation to express \( y \) in terms of \( x \):
\[
3y = x + 1 \implies y = \frac{x + 1}{3}
\]
Now we can use this equation to find \( B \) and \( A \).
**Finding \( B \):**
Using the point \( (−1, B) \):
\[
B = \frac{−1 + 1}{3} = \frac{0}{3} = 0
\]
So, we have:
\[
B = 0
\]
**Finding \( A \):**
Using the point \( (A, 2) \):
\[
2 = \frac{A + 1}{3}
\]
Now, we solve for \( A \):
Multiply both sides by \( 3 \):
\[
6 = A + 1
\]
Subtract \( 1 \) from both sides:
\[
A = 6 - 1 = 5
\]
So, we have:
\[
A = 5
\]
In summary, the values are:
\[
A = 5 \quad \text{and} \quad B = 0
\]
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