Question 27 1:28:20 is poorly worded. It should be worded as: “The equation has no real solutions if c>n. What is the value of n?” There is only one value of n here that is of interest, n = 289. When they ask the question as “least possible value”, they are implying that there are other higher values, they just want the least possible value. Although n = 290 has no real solutions, it would not be considered a correct answer because it does not cover the condition n = 289. For some reason, SAT wants to attach the term “least possible value” whenever they can.
Ha! Yeah that’s a brutal question… that and number 27 on that module… Basically, you have to find the slope of the first equation by putting it into y = mx + b form… once you have the slope, of the first, you can figure out the slope of the second (which we know is perpendicular) by taking the negative inverse of the original slope, which I explain the in video. Once you have slope of the second, perpendicular line, you can put the second equation into y = mx + b form to get your a and b values… Once you know your a and b values, you can plug those into the answer choices, graphing them in Desmos, to see which answer choice also gives you two perpendicular lines! That’s the easiest way to do it, although in the video I show you how to determine which ones are perpendicular algebraically… now I’m telling students just to Desmos it- by far the faster and easier method.
I'm binge watching all of your videos been 2 days and im done!!
We’ll done!! How are you feeling about the test?
@@MichaelToohey still not quite confident honestly! Do you think 2-3 weeks would be enough for me to touch a 1500
Question 27 1:28:20 is poorly worded. It should be worded as:
“The equation has no real solutions if c>n. What is the value of n?”
There is only one value of n here that is of interest, n = 289. When they ask the question as “least possible value”, they are implying that there are other higher values, they just want the least possible value.
Although n = 290 has no real solutions, it would not be considered a correct answer because it does not cover the condition n = 289. For some reason, SAT wants to attach the term “least possible value” whenever they can.
I agree! It’s very confusingly worded…
can you please explain the 26th question again
Ha! Yeah that’s a brutal question… that and number 27 on that module…
Basically, you have to find the slope of the first equation by putting it into y = mx + b form… once you have the slope, of the first, you can figure out the slope of the second (which we know is perpendicular) by taking the negative inverse of the original slope, which I explain the in video.
Once you have slope of the second, perpendicular line, you can put the second equation into y = mx + b form to get your a and b values…
Once you know your a and b values, you can plug those into the answer choices, graphing them in Desmos, to see which answer choice also gives you two perpendicular lines! That’s the easiest way to do it, although in the video I show you how to determine which ones are perpendicular algebraically… now I’m telling students just to Desmos it- by far the faster and easier method.
Make any sense?
we can also find out the slope by taking it in the form of -a/b right ?
thank you so much for taking your time to explain it .
@poornasahithi yes! That works! And that’s also what you end up with when you put the second equation in y = mx + b form…
@@MichaelToohey oh yes that makes sense . thank you so much!