SAT (Fall 2023)
- Solving quadratic equations — Basic example
- Solving quadratic equations — Harder example
- Interpreting nonlinear expressions — Basic example
- Interpreting nonlinear expressions — Harder example
- Quadratic and exponential word problems — Basic example
- Quadratic and exponential word problems — Harder example
- Manipulating quadratic and exponential expressions — Basic example
- Manipulating quadratic and exponential expressions — Harder example
- Radicals and rational exponents — Basic example
- Radicals and rational exponents — Harder example
- Radical and rational equations — Basic example
- Radical and rational equations — Harder example
- Operations with rational expressions — Basic example
- Operations with rational expressions — Harder example
- Operations with polynomials — Basic example
- Operations with polynomials — Harder example
- Polynomial factors and graphs — Basic example
- Polynomial factors and graphs — Harder example
- Nonlinear equation graphs — Basic example
- Nonlinear equation graphs — Harder example
- Linear and quadratic systems — Basic example
- Linear and quadratic systems — Harder example
- Structure in expressions — Basic example
- Structure in expressions — Harder example
- Isolating quantities — Basic example
- Isolating quantities — Harder example
- Function notation — Basic example
- Function notation — Harder example
Watch Sal work through a basic Manipulating quadratic and exponential expressions problem.
Want to join the conversation?
- why make things so much harder if all you have to do is plug in for the exponent of 2. you get the same answer of .7225(38 votes)
- Why make things more complicated? This could have been answered with common sense, 85% of 85% is going to be less than .85 so you pick A because that is the only one that is less than .85 and move on.(20 votes)
- Sometimes when your under a time crunch, you don't detect the simple tricks you could use and instead use methods you're more familiar with even if they're more complicated and time-consuming.(8 votes)
- This isn't a question. But a quicker way is:
when the last digit of 2 different values add up to 10, and the first digit(s) are the same, you can simply multiply the first digit(s) by the next digit(s) in succession. Then multiply the last digits by one another and put it behind the former.
85 * 85 //5 + 5 = 10; both have the same digit in the tens place so
9 * 8 = 72
24 * 26 = 624 // 6 + 4 = 10; 5 * 4 = 24
123 * 127 = 15621 // 12 * 13 = 156; 3 * 7 = 21(12 votes)
- If you put 2 in place of the variable that means you do this:
I am very confused, wouldn't it be be (D), because you also multiply by 24900.
Please help me out. I appreciate it.
Thank you(3 votes)
- Okay so option D says 1.7 right?? which isn't correct in this case...You are right, we multiply by 24900 to get the value two years later, but we also have to divide it by the current value since the question says "how many times its present value" and the present value is 24900, so it is effectively (24900*(0.85^2))/24900 which is just 0.85^2=0.7225(7 votes)
- Or...you could just multiply 0.85 by itself because after two years it will be 85% of 85%. It could be 30 seconds rather than 3 minutes.(5 votes)
- I think that all you people are talking about another lesson, rather than talking about what Sal is explaining.
I would say there some issues with that video, maybe it mix up with another video.(3 votes)
- You don't need to solve the question the answer is obvious by elimination(2 votes)
- Yes, you could do that. But would you want to do that if this was not on the calculator allowed side of the SAT?(2 votes)
- at0:22when t=0 and the problem is 0.85 to the zero power, how does that equal one? wouldn't it equal zero? I'm confused.(1 vote)
- [Instructor] We're asked, which of the following is an equivalent form of the equation of the graph shown in the xy-plane from which the x-intercepts can be identified as constants in the equation? So pause this video and see if you can figure this out. All right, now let's work through this together. So we wanna find an equation where if we graph it, we get this parabola right over here, and we want it to be in the form in which or from which the x-intercepts can be identified as constants in the equation. Well, let's look at the x-intercepts here. We have an x-intercept right over here. X-intercepts happen when y is equal to zero when we intersect the x axis. So one happens at x equals two, and then another here happens at x is equal to negative three. And so when we look at the choices, we are looking for a two or a three and the first two really don't see that. You aren't able to pick out these x-intercepts easily from choices A or B, so you can rule those out. Now, both choices C and D have some things that deal with threes and twos here. And what we have to remember is these are the x values that make y equal to zero. And so when you have it in this factored form, when you have this quadratic here in a factored form, if you have x minus two, that means that x equals two is going to be an x-intercept. How do we know that? Well, if you put a two in right over here, two minus two is zero, zero times anything is zero. If you put in a negative three here, negative three plus three is going to be zero. So this is actually the choice that we are looking at. Once you factor this quadratic in a form like this the intercepts are actually going to be the opposite, the negatives, of these numbers right over here. So this C right over here was a distractor to say, Oh look, I see a negative three and I see a two, but this is actually a different equation than what we have over here. And you would see here that the x-intercepts would be at x equals three and x equals negative two, not negative three and positive two, so we'd rule that one out as well.