Asked by 123
How do I solve for R?
1/[(1+R)^27]=0.423660792
1/[(1+R)^27]=0.423660792
Answers
Answered by
123
Is there any human tutor available because you don't know how to show people how to do math at all? I don't know who program you on math, but that person is dump at math. That why he program you to show steps that don't make sense at all.
Answered by
GPT-4o mini
To solve for \( R \) in the equation
\[
\frac{1}{(1+R)^{27}} = 0.423660792
\]
let's first rewrite the equation to isolate \( (1 + R)^{27} \):
1. Take the reciprocal of both sides:
\[
(1 + R)^{27} = \frac{1}{0.423660792}
\]
2. Calculate the right side:
\[
\frac{1}{0.423660792} \approx 2.36
\]
3. Now you have:
\[
(1 + R)^{27} \approx 2.36
\]
4. To solve for \( 1 + R \), take the 27th root of both sides:
\[
1 + R = (2.36)^{1/27}
\]
5. Now calculate \( (2.36)^{1/27} \):
Using a calculator or computational software, you can find:
\[
(2.36)^{1/27} \approx 1.034
\]
6. Now subtract 1 from both sides to solve for \( R \):
\[
R = (2.36)^{1/27} - 1
\]
Subtracting gives approximately:
\[
R \approx 1.034 - 1 \approx 0.034
\]
Thus, the value of \( R \) is about:
\[
R \approx 0.034 \text{ or } 3.4\%
\]
This means your solution for \( R \) is approximately \( 0.034 \) or \( 3.4\% \).
\[
\frac{1}{(1+R)^{27}} = 0.423660792
\]
let's first rewrite the equation to isolate \( (1 + R)^{27} \):
1. Take the reciprocal of both sides:
\[
(1 + R)^{27} = \frac{1}{0.423660792}
\]
2. Calculate the right side:
\[
\frac{1}{0.423660792} \approx 2.36
\]
3. Now you have:
\[
(1 + R)^{27} \approx 2.36
\]
4. To solve for \( 1 + R \), take the 27th root of both sides:
\[
1 + R = (2.36)^{1/27}
\]
5. Now calculate \( (2.36)^{1/27} \):
Using a calculator or computational software, you can find:
\[
(2.36)^{1/27} \approx 1.034
\]
6. Now subtract 1 from both sides to solve for \( R \):
\[
R = (2.36)^{1/27} - 1
\]
Subtracting gives approximately:
\[
R \approx 1.034 - 1 \approx 0.034
\]
Thus, the value of \( R \) is about:
\[
R \approx 0.034 \text{ or } 3.4\%
\]
This means your solution for \( R \) is approximately \( 0.034 \) or \( 3.4\% \).
Answered by
GPT-4o mini
I apologize for any confusion in my previous response. I'm here to help, and I'm committed to providing clear explanations. Let's clarify the solution step-by-step.
We started with the equation:
\[
\frac{1}{(1+R)^{27}} = 0.423660792
\]
### Step 1: Take the reciprocal
We can rearrange the equation:
\[
(1 + R)^{27} = \frac{1}{0.423660792}
\]
### Step 2: Calculate the right side
Calculating the right side gives:
\[
(1 + R)^{27} \approx 2.36 \quad (\text{since } \frac{1}{0.423660792} \approx 2.36)
\]
### Step 3: Take the 27th root
Now, we take the 27th root of both sides:
\[
1 + R = (2.36)^{1/27}
\]
### Step 4: Calculate \( (2.36)^{1/27} \)
This step requires either a calculator or logarithmic functions. Using a calculator:
\[
(2.36)^{1/27} \approx 1.0323 \quad (\text{approximately})
\]
### Step 5: Solve for \( R \)
Now, we subtract 1:
\[
R \approx 1.0323 - 1 \approx 0.0323
\]
### Final Answer
Thus, the value of \( R \) is approximately:
\[
R \approx 0.0323 \text{ or } 3.23\%
\]
I hope this clarification makes the solution clearer! If you have any more questions or need further assistance, just let me know.
We started with the equation:
\[
\frac{1}{(1+R)^{27}} = 0.423660792
\]
### Step 1: Take the reciprocal
We can rearrange the equation:
\[
(1 + R)^{27} = \frac{1}{0.423660792}
\]
### Step 2: Calculate the right side
Calculating the right side gives:
\[
(1 + R)^{27} \approx 2.36 \quad (\text{since } \frac{1}{0.423660792} \approx 2.36)
\]
### Step 3: Take the 27th root
Now, we take the 27th root of both sides:
\[
1 + R = (2.36)^{1/27}
\]
### Step 4: Calculate \( (2.36)^{1/27} \)
This step requires either a calculator or logarithmic functions. Using a calculator:
\[
(2.36)^{1/27} \approx 1.0323 \quad (\text{approximately})
\]
### Step 5: Solve for \( R \)
Now, we subtract 1:
\[
R \approx 1.0323 - 1 \approx 0.0323
\]
### Final Answer
Thus, the value of \( R \) is approximately:
\[
R \approx 0.0323 \text{ or } 3.23\%
\]
I hope this clarification makes the solution clearer! If you have any more questions or need further assistance, just let me know.
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