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Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve: \(\frac{1}{3}(x-6)+4 x>0\)

Short Answer

Expert verified
\( x > \frac{3}{13} \text{,or approximately x} > 0.462.

Step by step solution

01

Distribute the fraction

Distribute \(\frac{1}{3}\) through \((x - 6)\). This gives \(\frac{1}{3}x - 2\).
02

Combine like terms

Combine \(\frac{1}{3}x\) and \(+4x\). This simplifies to \( \frac{1}{3}x + 4x - 2 > 0 \).
03

Simplify the equation

Convert \(\frac{1}{3}x\) to decimals by finding a common denominator. \( \frac{1}{3}x\text{ is the same as }x/3 \). So, it can be rewritten as: \( \frac{1}{3}x = 0.3333x \text{ or } 0.3333x \). Combine with \(4x\) to get: \( \0.3333x + 4x \).
04

Solve for x

Combine terms and solve for x: \(4.3333x > 2 \). Divide both sides by 4.3333: \(\frac{2}{4.3333}= x \). This simplifies to \(\frac{2}{4.33} \text{is approximately equal to } x \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving linear inequalities
Solving linear inequalities is a fundamental aspect of algebra. It involves finding the values of a variable that satisfy the given inequality. Here, we are tasked with solving \(\frac{1}{3}(x-6)+4x>0\).

To solve this, follow these steps:
  • First, we distribute any fractions or coefficients through parentheses.

  • Next, combine all like terms to simplify the expression.

  • Finally, solve for the variable by isolating it on one side of the inequality.

Notice that solving an inequality is quite similar to solving an equation, but we must remember the crucial rule: if we multiply or divide both sides by a negative number, we must reverse the inequality sign.
combining like terms
Combining like terms allows us to simplify expressions more effectively. Like terms are terms that contain the same variable raised to the same power.

In the inequality \(\frac{1}{3}(x-6)+4x>0\), we distribute \(\frac{1}{3}\) through each term to get \(\frac{1}{3}x - 2\).

Now we have: \(\frac{1}{3}x - 2 + 4x > 0\).
Combine the like terms \(\frac{1}{3}x\) and \(4x\).
Converting \(\frac{1}{3}x\) to decimal form, we get 0.3333x, so we add 0.3333x and 4x:

\(\frac{1}{3}x + 4x \rightarrow 0.3333x + 4x = 4.3333x\). Now, our inequality looks like this: 4.3333x - 2 > 0. By combining like terms, we effectively reduce the complexity of the expression, making it easier to solve.
simplifying expressions
Simplifying expressions involves making them as concise as possible without changing their value. This often involves combining like terms and performing arithmetic operations.

Let's consider the expression we derived: 4.3333x - 2 > 0.

Now, we need to isolate x.
  • Add 2 to both sides to get 4.3333x > 2.

  • Then, divide both sides by 4.3333 to isolate x.

This gives us:
\ x > \frac{2}{4.3333} \
Approximating, we get \ x > \frac{2}{4.33} = 0.461\

Hence, the solution to the inequality is \( x > 0.461 \). Simplifying expressions is crucial for making calculations more manageable, especially when dealing with inequalities.

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Most popular questions from this chapter

In Problems \(47-52,\) the function \(d\) models the distance (in meters) of the bob of a pendulum of mass \(m\) (in kilograms) from its rest position at time \(t\) (in seconds) is given. The bob is released from the left of its rest position and represents a negative direction. (a) Describe the motion of the object. Be sure to give the mass and damping factor. (b) What is the initial displacement of the bob? That is, what is the displacement at \(t=0 ?\) (c) Graph the motion using a graphing utility. (d) What is the displacement of the bob at the start of the second oscillation? (e) What happens to the displacement of the bob as time increases without bound? $$ d(t)=-20 e^{-0.7 t / 40} \cos \left(\sqrt{\left(\frac{2 \pi}{5}\right)^{2}-\frac{0.49}{1600}} t\right) $$

The function \(d\) models the distance (in meters) of the bob of a pendulum of mass \(m\) (in kilograms) from its rest position at time \(t\) (in seconds) is given. The bob is released from the left of its rest position and represents a negative direction. (a) Describe the motion of the object. Be sure to give the mass and damping factor. (b) What is the initial displacement of the bob? That is, what is the displacement at \(t=0 ?\) (c) Graph the motion using a graphing utility. (d) What is the displacement of the bob at the start of the second oscillation? (e) What happens to the displacement of the bob as time increases without bound? $$ d(t)=-30 e^{-0.6 t / 80} \cos \left(\sqrt{\left(\frac{2 \pi}{7}\right)^{2}-\frac{0.36}{6400}} t\right) $$

(a) use the Product-to-Sum Formulas to express each product as a sum, and (b) use the method of adding \(y\) -coordinates to graph each function on the interval \([0,2 \pi] .\) $$ h(x)=\cos (2 x) \cos (x) $$

An object of mass \(m\) (in grams) attached to a coiled spring with damping factor \(b\) (in grams per second) is pulled down a distance a (in centimeters) from its rest position and then released. Assume that the positive direction of the motion is up and the period is \(T\) (in seconds) under simple harmonic motion. (a) Find a function that relates the displacement d of the object from its rest position after \(t\) seconds. (b) Graph the function found in part (a) for 5 oscillations using a graphing utility. $$ m=20, \quad a=15, \quad b=0.75, \quad T=6 $$

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve \(4^{x}=3^{x+1}\). Express the solution in exact form.

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