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Original Question
Use the graph to answer the question. Write the exponential equation shown in the graph. A coordinate plane going from negative...Question
Use the graph to answer the question.
Write the exponential equation shown in the graph.
A coordinate plane going from negative 5 to 5 in increments of 1 labeled every 5 units on the x-axis and from negative 1 to 10 in increments of 1 labeled every 5 units on the y-axis.
there are two makred points, 1,3 and 0,1 and the arrow is horizontal at first then it curves into the upper fight quadrant from the upper left.
(1 point)
The equation is $$
.
Write the exponential equation shown in the graph.
A coordinate plane going from negative 5 to 5 in increments of 1 labeled every 5 units on the x-axis and from negative 1 to 10 in increments of 1 labeled every 5 units on the y-axis.
there are two makred points, 1,3 and 0,1 and the arrow is horizontal at first then it curves into the upper fight quadrant from the upper left.
(1 point)
The equation is $$
.
Answers
Answer
Use the graph to answer the question.
Write the exponential equation shown in the graph.
A coordinate plane going from negative 5 to 5 in increments of 1 labeled every 5 units on the x-axis and from negative 1 to 10 in increments of 1 labeled every 5 units on the y-axis.
the marked points are - 1, 1/3 and 1, 3/4
(1 point)
The equation is $$
.
Write the exponential equation shown in the graph.
A coordinate plane going from negative 5 to 5 in increments of 1 labeled every 5 units on the x-axis and from negative 1 to 10 in increments of 1 labeled every 5 units on the y-axis.
the marked points are - 1, 1/3 and 1, 3/4
(1 point)
The equation is $$
.
Answer
Question
The population of certain insects is initially 312. The insect population shows exponential growth of 5% each week. Write the exponential equation to model the insect population, y, after x weeks.
The population of certain insects is initially 312. The insect population shows exponential growth of 5% each week. Write the exponential equation to model the insect population, y, after x weeks.
Answer
Write the exponential equation that contains the points (−2,4)
and (3,972)
.(1 point)
The equation is $$
.
and (3,972)
.(1 point)
The equation is $$
.
Answered by
GPT-4o mini
To write the equation of the exponential function based on the description you've provided, we can use the general form of an exponential function:
\[
y = ab^x
\]
where:
- \(a\) is the initial value (the value of \(y\) when \(x=0\)),
- \(b\) is the base of the exponential which determines the growth rate.
From the points you provided:
1. The point (0, 1) indicates that when \(x = 0\), \(y = 1\). This means \(a = 1\).
2. The point (1, 3) shows that when \(x = 1\), \(y = 3\).
So, we can substitute the first point into the exponential equation:
\[
1 = ab^0 \quad \Rightarrow \quad a = 1
\]
Now, we substitute the second point into the equation to find \(b\):
\[
3 = 1 \cdot b^1 \quad \Rightarrow \quad b = 3
\]
Thus, the equation of the exponential function is:
\[
y = 1 \cdot 3^x
\]
This simplifies to:
\[
y = 3^x
\]
So, the final answer is:
\[
\text{The equation is } y = 3^x.
\]
\[
y = ab^x
\]
where:
- \(a\) is the initial value (the value of \(y\) when \(x=0\)),
- \(b\) is the base of the exponential which determines the growth rate.
From the points you provided:
1. The point (0, 1) indicates that when \(x = 0\), \(y = 1\). This means \(a = 1\).
2. The point (1, 3) shows that when \(x = 1\), \(y = 3\).
So, we can substitute the first point into the exponential equation:
\[
1 = ab^0 \quad \Rightarrow \quad a = 1
\]
Now, we substitute the second point into the equation to find \(b\):
\[
3 = 1 \cdot b^1 \quad \Rightarrow \quad b = 3
\]
Thus, the equation of the exponential function is:
\[
y = 1 \cdot 3^x
\]
This simplifies to:
\[
y = 3^x
\]
So, the final answer is:
\[
\text{The equation is } y = 3^x.
\]
Answered by
GPT-4o mini
To find the exponential equation based on the points you provided, we can again use the general form of an exponential function:
\[
y = ab^x
\]
where:
- \(a\) is the initial value (the value of \(y\) when \(x=0\)),
- \(b\) is the base of the exponential function.
### Given Points:
1. The point \((-1, \frac{1}{3})\)
2. The point \((1, \frac{3}{4})\)
### Step 1: Finding \(a\)
To find \(a\), we can assume the function passes through the point \((0, a)\). However, since we don't have the value directly, we will solve for \(a\) later.
### Step 2: Finding \(b\) using the two points
We can write the equations corresponding to each of these points:
1. For the point \((-1, \frac{1}{3})\):
\[
\frac{1}{3} = ab^{-1} \quad \Rightarrow \quad \frac{1}{3} = \frac{a}{b}
\]
Thus, we rearrange this to get:
\[
a = \frac{1}{3} b
\]
2. For the point \((1, \frac{3}{4})\):
\[
\frac{3}{4} = ab^{1} \quad \Rightarrow \quad \frac{3}{4} = ab
\]
### Step 3: Substitute \(a\) into the second equation
Now substitute \(a = \frac{1}{3} b\) into the second equation:
\[
\frac{3}{4} = \left(\frac{1}{3} b\right)b
\]
This simplifies to:
\[
\frac{3}{4} = \frac{1}{3} b^2
\]
### Step 4: Solve for \(b^2\)
Multiply both sides by 3:
\[
\frac{9}{4} = b^2
\]
Now take the square root of both sides:
\[
b = \frac{3}{2} \quad \text{(since \(b\) must be positive)}
\]
### Step 5: Find \(a\)
Now we can substitute \(b = \frac{3}{2}\) back into the equation for \(a\):
\[
a = \frac{1}{3} b = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2}
\]
### Final Equation
Now we have \(a\) and \(b\):
- \(a = \frac{1}{2}\)
- \(b = \frac{3}{2}\)
Thus, the exponential equation is:
\[
y = \frac{1}{2} \left(\frac{3}{2}\right)^x
\]
So the final answer is:
\[
\text{The equation is } y = \frac{1}{2} \left(\frac{3}{2}\right)^x.
\]
\[
y = ab^x
\]
where:
- \(a\) is the initial value (the value of \(y\) when \(x=0\)),
- \(b\) is the base of the exponential function.
### Given Points:
1. The point \((-1, \frac{1}{3})\)
2. The point \((1, \frac{3}{4})\)
### Step 1: Finding \(a\)
To find \(a\), we can assume the function passes through the point \((0, a)\). However, since we don't have the value directly, we will solve for \(a\) later.
### Step 2: Finding \(b\) using the two points
We can write the equations corresponding to each of these points:
1. For the point \((-1, \frac{1}{3})\):
\[
\frac{1}{3} = ab^{-1} \quad \Rightarrow \quad \frac{1}{3} = \frac{a}{b}
\]
Thus, we rearrange this to get:
\[
a = \frac{1}{3} b
\]
2. For the point \((1, \frac{3}{4})\):
\[
\frac{3}{4} = ab^{1} \quad \Rightarrow \quad \frac{3}{4} = ab
\]
### Step 3: Substitute \(a\) into the second equation
Now substitute \(a = \frac{1}{3} b\) into the second equation:
\[
\frac{3}{4} = \left(\frac{1}{3} b\right)b
\]
This simplifies to:
\[
\frac{3}{4} = \frac{1}{3} b^2
\]
### Step 4: Solve for \(b^2\)
Multiply both sides by 3:
\[
\frac{9}{4} = b^2
\]
Now take the square root of both sides:
\[
b = \frac{3}{2} \quad \text{(since \(b\) must be positive)}
\]
### Step 5: Find \(a\)
Now we can substitute \(b = \frac{3}{2}\) back into the equation for \(a\):
\[
a = \frac{1}{3} b = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2}
\]
### Final Equation
Now we have \(a\) and \(b\):
- \(a = \frac{1}{2}\)
- \(b = \frac{3}{2}\)
Thus, the exponential equation is:
\[
y = \frac{1}{2} \left(\frac{3}{2}\right)^x
\]
So the final answer is:
\[
\text{The equation is } y = \frac{1}{2} \left(\frac{3}{2}\right)^x.
\]
Answered by
GPT-4o mini
To model the exponential growth of the insect population, we can use the general formula for exponential growth, which is:
\[
y = a(1 + r)^x
\]
where:
- \(y\) is the population after \(x\) weeks,
- \(a\) is the initial population,
- \(r\) is the growth rate (as a decimal),
- \(x\) is the number of time periods (weeks in this case).
### Given Data:
- The initial population \(a = 312\),
- The growth rate \(r = 5\% = 0.05\).
### Step 1: Substitute the values into the equation
Substituting \(a\) and \(r\) into the exponential growth formula, we get:
\[
y = 312(1 + 0.05)^x
\]
which simplifies to:
\[
y = 312(1.05)^x
\]
### Final Equation
Thus, the exponential equation that models the insect population after \(x\) weeks is:
\[
y = 312(1.05)^x
\]
\[
y = a(1 + r)^x
\]
where:
- \(y\) is the population after \(x\) weeks,
- \(a\) is the initial population,
- \(r\) is the growth rate (as a decimal),
- \(x\) is the number of time periods (weeks in this case).
### Given Data:
- The initial population \(a = 312\),
- The growth rate \(r = 5\% = 0.05\).
### Step 1: Substitute the values into the equation
Substituting \(a\) and \(r\) into the exponential growth formula, we get:
\[
y = 312(1 + 0.05)^x
\]
which simplifies to:
\[
y = 312(1.05)^x
\]
### Final Equation
Thus, the exponential equation that models the insect population after \(x\) weeks is:
\[
y = 312(1.05)^x
\]
Answered by
GPT-4o mini
To find the exponential equation that passes through the points \((-2, 4)\) and \((3, 972)\), we can use the general form of an exponential function:
\[
y = ab^x
\]
where:
- \(a\) is the initial value (or value when \(x = 0\)),
- \(b\) is the base of the exponential function.
Given the two points, we can set up two equations:
1. For the point \((-2, 4)\):
\[
4 = ab^{-2} \quad \Rightarrow \quad 4 = \frac{a}{b^2}
\]
This can be rearranged as:
\[
a = 4b^2 \quad \quad \text{(Equation 1)}
\]
2. For the point \((3, 972)\):
\[
972 = ab^3 \quad \Rightarrow \quad 972 = a \cdot b^3 \quad \quad \text{(Equation 2)}
\]
### Step 1: Substitute Equation 1 into Equation 2
Substituting \(a = 4b^2\) from Equation 1 into Equation 2:
\[
972 = (4b^2) b^3
\]
This simplifies to:
\[
972 = 4b^{5}
\]
### Step 2: Solve for \(b^5\)
Now, divide both sides by 4:
\[
b^{5} = \frac{972}{4} = 243
\]
### Step 3: Solve for \(b\)
Next, take the fifth root of both sides:
\[
b = 243^{\frac{1}{5}}
\]
To find \(243\), we note that \(243 = 3^5\), so:
\[
b = 3
\]
### Step 4: Substitute \(b\) back to find \(a\)
Now substitute \(b = 3\) back into Equation 1:
\[
a = 4b^2 = 4(3^2) = 4 \cdot 9 = 36
\]
### Final Exponential Equation
Now we have \(a\) and \(b\):
\[
a = 36,\quad b = 3
\]
Thus, the exponential equation is:
\[
y = 36 \cdot 3^x
\]
So, the final answer is:
\[
\text{The equation is } y = 36 \cdot 3^x.
\]
\[
y = ab^x
\]
where:
- \(a\) is the initial value (or value when \(x = 0\)),
- \(b\) is the base of the exponential function.
Given the two points, we can set up two equations:
1. For the point \((-2, 4)\):
\[
4 = ab^{-2} \quad \Rightarrow \quad 4 = \frac{a}{b^2}
\]
This can be rearranged as:
\[
a = 4b^2 \quad \quad \text{(Equation 1)}
\]
2. For the point \((3, 972)\):
\[
972 = ab^3 \quad \Rightarrow \quad 972 = a \cdot b^3 \quad \quad \text{(Equation 2)}
\]
### Step 1: Substitute Equation 1 into Equation 2
Substituting \(a = 4b^2\) from Equation 1 into Equation 2:
\[
972 = (4b^2) b^3
\]
This simplifies to:
\[
972 = 4b^{5}
\]
### Step 2: Solve for \(b^5\)
Now, divide both sides by 4:
\[
b^{5} = \frac{972}{4} = 243
\]
### Step 3: Solve for \(b\)
Next, take the fifth root of both sides:
\[
b = 243^{\frac{1}{5}}
\]
To find \(243\), we note that \(243 = 3^5\), so:
\[
b = 3
\]
### Step 4: Substitute \(b\) back to find \(a\)
Now substitute \(b = 3\) back into Equation 1:
\[
a = 4b^2 = 4(3^2) = 4 \cdot 9 = 36
\]
### Final Exponential Equation
Now we have \(a\) and \(b\):
\[
a = 36,\quad b = 3
\]
Thus, the exponential equation is:
\[
y = 36 \cdot 3^x
\]
So, the final answer is:
\[
\text{The equation is } y = 36 \cdot 3^x.
\]
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