Problem 7
Question
For Problems \(1-10\), solve for the specified variable using the given facts. (Objective 1) $$ \text { Solve } V=\frac{1}{3} B h \quad \text { for } B \text { if } V=112 \text { and } h=7 $$
Step-by-Step Solution
Verified Answer
The value of \( B \) is approximately 48.
1Step 1: Substitute Known Values
Start by substituting the given values for \( V \) and \( h \) into the formula. We have \( V = 112 \) and \( h = 7 \). So the equation becomes:\[ 112 = \frac{1}{3} B \times 7 \]
2Step 2: Simplify the Equation
Simplify the equation by multiplying the fraction by 7:\[ 112 = \frac{7}{3} B \]
3Step 3: Solve for \( B \)
To isolate \( B \), multiply both sides of the equation by \( \frac{3}{7} \):\[ B = 112 \times \frac{3}{7} \]
4Step 4: Calculate \( B \)
Perform the multiplication:\[ B = 112 \times \frac{3}{7} = 112 \times 0.4286 \approx 48 \]
Key Concepts
AlgebraVariable IsolationMathematical Substitution
Algebra
Algebra is a branch of mathematics that helps us solve equations involving variables or unknown values. The goal of algebra is to determine the values of these unknowns by using known quantities and mathematical operations.
In algebra, we often deal with formulas and need to manipulate them to solve for a particular variable. These formulas can describe real-world situations, like the volume of a shape, the distance traveled, or even the speed of an object. By substituting known values into these formulas, we can figure out unknown ones.
Algebra uses various operations such as addition, subtraction, multiplication, and division to manipulate equations and isolate variables. This manipulation helps us understand relationships between different quantities and solve practical problems. Remember, algebra is not just about numbers; it’s about understanding patterns and logical thinking.
In algebra, we often deal with formulas and need to manipulate them to solve for a particular variable. These formulas can describe real-world situations, like the volume of a shape, the distance traveled, or even the speed of an object. By substituting known values into these formulas, we can figure out unknown ones.
Algebra uses various operations such as addition, subtraction, multiplication, and division to manipulate equations and isolate variables. This manipulation helps us understand relationships between different quantities and solve practical problems. Remember, algebra is not just about numbers; it’s about understanding patterns and logical thinking.
Variable Isolation
Variable isolation is crucial in solving equations. It involves rearranging equations to make one variable stand alone.
To isolate a variable, we need to perform inverse operations that undo what's currently being done to the variable. If the variable is being multiplied by a number, you divide both sides by that number. If it's being added to, subtract it from both sides, and vice versa. This step-by-step process allows you to bring the variable to one side of the equation without changing the equality.
For example, in the exercise provided, to find the value of 'B', we worked on isolating 'B' from the equation. We had to remove the fractions and operations around 'B'. Each step made 'B' more accessible until it stood alone on one side of the equation. Practicing this helps build a strong foundation in solving algebraic equations.
To isolate a variable, we need to perform inverse operations that undo what's currently being done to the variable. If the variable is being multiplied by a number, you divide both sides by that number. If it's being added to, subtract it from both sides, and vice versa. This step-by-step process allows you to bring the variable to one side of the equation without changing the equality.
For example, in the exercise provided, to find the value of 'B', we worked on isolating 'B' from the equation. We had to remove the fractions and operations around 'B'. Each step made 'B' more accessible until it stood alone on one side of the equation. Practicing this helps build a strong foundation in solving algebraic equations.
Mathematical Substitution
Mathematical substitution is the process of replacing variables with known values. This is useful when an equation involves more than one variable but you need to solve for just one.
In our exercise, we knew 'V' and 'h' and had to find 'B'. The first step was to substitute these known values into the equation to simplify it. By replacing 'V=112' and 'h=7', the equation becomes more manageable.
This concept shows the power of substitution in algebra, making complex equations simpler. By inserting actual numbers, we transform an abstract equation into an arithmetic problem, which is generally easier to solve. Substitution is a key tool for making algebra practical, letting us plug real-world data into mathematical formulas to find solutions.
In our exercise, we knew 'V' and 'h' and had to find 'B'. The first step was to substitute these known values into the equation to simplify it. By replacing 'V=112' and 'h=7', the equation becomes more manageable.
This concept shows the power of substitution in algebra, making complex equations simpler. By inserting actual numbers, we transform an abstract equation into an arithmetic problem, which is generally easier to solve. Substitution is a key tool for making algebra practical, letting us plug real-world data into mathematical formulas to find solutions.
Other exercises in this chapter
Problem 7
For Problems 1-12, solve each equation. You will be using these types of equations in Problems \(13-41\). $$ 0.4(10)-0.4 x+x=0.5(10) $$
View solution Problem 7
For Problems \(1-12\), solve each of the equations. These equations are the types you will be using in Problems 13-40. $$ s+(2 s-1)+(3 s-4)=37 $$
View solution Problem 7
Solve each of the equations. $$0.7 r=56$$
View solution Problem 7
Solve each of the equations. $$\frac{x-2}{4}=\frac{x+4}{3}$$
View solution