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00:55

Taylor S.

Using Models Use the model given to answer the questions about the object or process being modeled. The volume $V$ of a cylindrical can is modeled by the formula $$V=\pi r^{2} h$$ where $r$ is the radius and $h$ is the height of the can. Find the volume of a can with radius 3 in. and height 5 in.

03:34

Kyle I.

Using Models Use the model given to answer the questions about the object or process being modeled. The distance $d$ (in mi) driven by a car traveling at a speed of $v$ miles per hour for $t$ hours is given by $$d=v t$$ If the car is driven at 70 milh for 3.5 $\mathrm{h}$ , how far has it traveled?

01:56

Suppose gas costs $\$ 3.50$ a gallon. We make a model for the cost $C$ of buying $x$ gallons of gas by writing the formula $C=$ _________.

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absolute value equation three minus the absolute value of eight X minus six equals three. Now we need to get the absolute value of eight X minus six on the left side by itself before we can even begin looking at solving. So we need to go ahead. And the first thing we're gonna do is we're gonna subtract three from both sites, so that's gonna cancel out. Leave me with negative Absolute value of eight X minus six equals zero. We've still got this negative in front of the absolute value, son. So let's go ahead and let's divide that by negative one. Now that's going to remove are negative and leave me with the absolute value of eight X minus six. Now, when I divide zero by negative one, I'm still gonna have zero. So that's not gonna change. So that means that I'm gonna work eight x minus six. Absolute value of eight X minus six equals zero Now, typically, when you're dealing with absolute value, you're gonna work it twice. So the first time I would work is eight x minus six equals zero. I would add six eight x equals six, divide by eight and I can simplify that. Teoh X equals 3/4. Then the second time I work, it is when I would change the number on the outside of the absolute value sign. So I still have eight. X minus six equals. Wait a second. I can't change zero to negative, because zero is neither negative or positive. So what I have is the exact same thing that I had the first time I worked it. So instead of working, get twice what you could and what you're gonna come up with this three force, you're just gonna have one solution to this problem, and your solution is three force.

Linear Equations and Functions

Linear Equations and Inequalities

Matrices and Determinants

Quadratic Equations

Applications of Trigonometric Functions

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