Solve 56 =7v, where v is a real number. Round your answer to the nearest hundredth. If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". V = 0.. No solution
Solve 56 =7v, where v is a real number. Round your answer to the nearest hundredth. If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". V = 0.. No solution
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![**Solving a Quadratic Equation Using the Square Root Property**
**Problem:**
Solve \( 56 = 7v^2 \), where \( v \) is a real number. Round your answer to the nearest hundredth.
**Instructions:**
1. Solve the given quadratic equation for \( v \).
2. If there is more than one solution, separate them with commas.
3. If there is no solution, click on "No solution".
**Options:**
- Input your answer into the text box labeled \( v = \).
- Click on "No solution" if applicable.
**Example:**
To solve the equation \( 56 = 7v^2 \):
1. Divide both sides of the equation by 7 to get \( v^2 \):
\[
v^2 = \frac{56}{7} = 8
\]
2. Take the square root of both sides, remembering to consider both the positive and negative square roots:
\[
v = \pm \sqrt{8} = \pm 2.83 \quad (\text{rounded to the nearest hundredth})
\]
3. The solutions are:
\[
v = 2.83, -2.83
\]
**Interface Features:**
- A text box for inputting the value(s) of \( v \).
- A "No solution" button to click if the equation has no real solution.
**Explanation and Check:**
- **Explanation Button:** Provides a detailed step-by-step solution to the problem.
- **Check Button:** Allows you to submit your answer to see if it is correct.
**Note:**
This example illustrates how to use the square root property to solve quadratic equations. For complex problems involving different coefficients or additional terms, additional algebraic steps may be required.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce668d18-9889-43e2-93d7-4b1cfcddd4ed%2F45084778-c0c2-4c82-adaf-1e21baa28a46%2Fo0riqum_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving a Quadratic Equation Using the Square Root Property**
**Problem:**
Solve \( 56 = 7v^2 \), where \( v \) is a real number. Round your answer to the nearest hundredth.
**Instructions:**
1. Solve the given quadratic equation for \( v \).
2. If there is more than one solution, separate them with commas.
3. If there is no solution, click on "No solution".
**Options:**
- Input your answer into the text box labeled \( v = \).
- Click on "No solution" if applicable.
**Example:**
To solve the equation \( 56 = 7v^2 \):
1. Divide both sides of the equation by 7 to get \( v^2 \):
\[
v^2 = \frac{56}{7} = 8
\]
2. Take the square root of both sides, remembering to consider both the positive and negative square roots:
\[
v = \pm \sqrt{8} = \pm 2.83 \quad (\text{rounded to the nearest hundredth})
\]
3. The solutions are:
\[
v = 2.83, -2.83
\]
**Interface Features:**
- A text box for inputting the value(s) of \( v \).
- A "No solution" button to click if the equation has no real solution.
**Explanation and Check:**
- **Explanation Button:** Provides a detailed step-by-step solution to the problem.
- **Check Button:** Allows you to submit your answer to see if it is correct.
**Note:**
This example illustrates how to use the square root property to solve quadratic equations. For complex problems involving different coefficients or additional terms, additional algebraic steps may be required.
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