The last question can be solved even simpler. Take the equation, plug in desmos. Find x values. Remember when u set the solution = 0, k will be positive. so set the positive solution I think it was 0.414 when I graph to the sqrt k - 1 and solve for k. Dont need complex math
You can’t compare the sat to the Jee the sat is a lot faster paced test and also has an English section that test your vocabulary knowledge of a couple 1000 words that aren’t like common sense words. And sat is for general admission even stuff like art schools not strictly for getting into prestigious engineering or math related programs. Even with a perfect sat your not garenteed a spot at selective schools.
Then, Why are you here for? Go and do your JEE shit . Giving SAT is far better than giving yourself a mental torture or studying like a slave for JEE . USA is far better option than staying in India after JEE clear.
What is the point of a comparison? It doesn’t make you smarter, plus these are 2 different tests. Jee is known to be extremely hard and the SAT is just a random standardized test in America not stared to be as hard as Jee, so idk why I see every Indian saying this.
@@zahin266 Even a non-math background 14 year old Indian student will find it extremely easy. Infact you'll find most of the 13-15 year old Indian kids bragging about how easy SAT is.
@@AmrutaVelamuri yea or just do it with your regular calculator, and log base 2/ 3 it. idk why he made it seem a bit too complicated not like we have a non calc section
@@senshinra First solve for B with the first equation, set b as x in this context and solve. After that do (2/3)^( x)=the b value you have. This way you dont need to actually any of the calculations
was waiting for this video. Waiting for more
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By doing what from the previous exponents got you 15/2 as x (like from 5/2 and 1/3).
you can use log (2/3) (b), which makes x the subject
The last question can be solved even simpler. Take the equation, plug in desmos. Find x values. Remember when u set the solution = 0, k will be positive. so set the positive solution I think it was 0.414 when I graph to the sqrt k - 1 and solve for k. Dont need complex math
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i'm confused, would the third question not be solved using discriminants? because it says No solution?
it is not a quadratic equation. It is linear
Didnt you teach at my school?
Im so confused with the first question
Bro as a jee aspirant it is nothing
You can’t compare the sat to the Jee the sat is a lot faster paced test and also has an English section that test your vocabulary knowledge of a couple 1000 words that aren’t like common sense words. And sat is for general admission even stuff like art schools not strictly for getting into prestigious engineering or math related programs. Even with a perfect sat your not garenteed a spot at selective schools.
Then, Why are you here for? Go and do your JEE shit . Giving SAT is far better than giving yourself a mental torture or studying like a slave for JEE . USA is far better option than staying in India after JEE clear.
What is the point of a comparison? It doesn’t make you smarter, plus these are 2 different tests. Jee is known to be extremely hard and the SAT is just a random standardized test in America not stared to be as hard as Jee, so idk why I see every Indian saying this.
@@zahin266 Even a non-math background 14 year old Indian student will find it extremely easy. Infact you'll find most of the 13-15 year old Indian kids bragging about how easy SAT is.
@@siddharthtiwary unfortunately that still does not answer my question for why a comparison is needed. These are, 2 way different tests
The first question could be done in an easier way if you know about logs
honestly plug it into desmos, way easier and less icky math
@@AmrutaVelamuri yea or just do it with your regular calculator, and log base 2/ 3 it. idk why he made it seem a bit too complicated not like we have a non calc section
@@AmrutaVelamuri what do you plug in?
@@nirmalkusunuri4240 how
@@senshinra First solve for B with the first equation, set b as x in this context and solve. After that do (2/3)^( x)=the b value you have. This way you dont need to actually any of the calculations