For the equation at 2:24 you most certainly can nail that with desmos as long as you reduce the step size (or make the interval very narrow) and zoom in hard. c=17.07 and this took no time, and it's accurate to the 5 symbols the SAT requires on student response questions.
The second problem (radicals) is ridiculously easy in desmos and took literally 10 seconds: enter the two equations and first check the bottom intersection point. Notice it's (-, -), so it can be ruled out of hand since none of the answers are (-, -). The top intersection is (+,+) which narrows it to A or C. Then just type in sqrt(3) to see if it matches the y-coordinate of 1.732.... It does, so A is the answer.
The first problem (ugly fractions) can most definitely be solved in desmos in 4 steps: (1) create a table with the points (0,7) and (8,0) given as clean intercepts, (2) enter the linear regression y1~mx1 + b, which hands you m (=-0.875) and b (=7) , (3) type 4 ~ -0.875d + 7 to get the answer for d (=3.42857), and (4) type d and click the fraction icon to convert to a fraction, 24/7. This took 30 seconds.
question 2 and 3 are easily solvable with desmos. For question 2 you can plug in both equations, find the intersection, then plug in the radicals to check which one is correct. For question 3, after you find out it is between 17 and 17.3, you can tighten the boundaries of the slider to that and find the lowest value of c that can make it one solution, that being 17.067
(quick note, I currently have 700 in Math, but March SAT was my first one, so I hope it will improve in the future) In the 3rd example (15x² - 32x + c = y), yes it is not efficient to plug the whole equation in desmos, but it is still useful to put 32² - 4(15)c = 0 because it reduces time and chance of error. Also, 256/15 is 6 characters long and 17.067 can be rounded to 17.07.
For 2, you can graph the two equations and see where they intersect. You can also estimate the solution as root of 3 is less than 2 and root of 5 is bigger than 2, so it becomes obvious in the grided diagram. 0 computation involved.
The last question is a joke and you don't even have to graph it. Just type in the expression, then immediately after it type ON THE SAME LINE 'with x = 4'. Desmos will evaluate the expression and give you the value 3 in this case. Note that you can use literally any number for x, except 3 (division by 0). Then type each answer using 'with x=4' after it and whichever matches the value of 3 is the answer. The answers are nearly the same, so it takes almost no effort to modify each one rather than type them from scratch. This took all of 15 seconds.
![[March SAT Math] These 3 Questions were JUST added](http://i.ytimg.com/vi/37YoPXLGAk4/mqdefault.jpg)
For the equation at 2:24 you most certainly can nail that with desmos as long as you reduce the step size (or make the interval very narrow) and zoom in hard. c=17.07 and this took no time, and it's accurate to the 5 symbols the SAT requires on student response questions.
17.067 is more appropriate than 17.07
The second problem (radicals) is ridiculously easy in desmos and took literally 10 seconds: enter the two equations and first check the bottom intersection point. Notice it's (-, -), so it can be ruled out of hand since none of the answers are (-, -). The top intersection is (+,+) which narrows it to A or C. Then just type in sqrt(3) to see if it matches the y-coordinate of 1.732.... It does, so A is the answer.
Love ur viveos! they are so helpful! thank you!
The first problem (ugly fractions) can most definitely be solved in desmos in 4 steps: (1) create a table with the points (0,7) and (8,0) given as clean intercepts, (2) enter the linear regression y1~mx1 + b, which hands you m (=-0.875) and b (=7) , (3) type 4 ~ -0.875d + 7 to get the answer for d (=3.42857), and (4) type d and click the fraction icon to convert to a fraction, 24/7. This took 30 seconds.
its answer is 24/7 ?
question 2 and 3 are easily solvable with desmos. For question 2 you can plug in both equations, find the intersection, then plug in the radicals to check which one is correct. For question 3, after you find out it is between 17 and 17.3, you can tighten the boundaries of the slider to that and find the lowest value of c that can make it one solution, that being 17.067
how do i tighten the boundaries
@@umon7751 if you click on either of the numbers on the slider it will show an inequality where you can pick different numbers
(quick note, I currently have 700 in Math, but March SAT was my first one, so I hope it will improve in the future)
In the 3rd example (15x² - 32x + c = y), yes it is not efficient to plug the whole equation in desmos, but it is still useful to put 32² - 4(15)c = 0 because it reduces time and chance of error. Also, 256/15 is 6 characters long and 17.067 can be rounded to 17.07.
For 2, you can graph the two equations and see where they intersect. You can also estimate the solution as root of 3 is less than 2 and root of 5 is bigger than 2, so it becomes obvious in the grided diagram. 0 computation involved.
Exactly. Plus, one of the solutions has a negative x & y coordinate so you can eliminate B & D right away and you only have to check 2 answers, not 4.
Title should be renamed: PERFECT DESMOS OPPORTUNITIES FOR THE DSAT!
The last question is a joke and you don't even have to graph it. Just type in the expression, then immediately after it type ON THE SAME LINE 'with x = 4'. Desmos will evaluate the expression and give you the value 3 in this case. Note that you can use literally any number for x, except 3 (division by 0). Then type each answer using 'with x=4' after it and whichever matches the value of 3 is the answer. The answers are nearly the same, so it takes almost no effort to modify each one rather than type them from scratch. This took all of 15 seconds.