SAT Khan Academy Graphing Linear Equations

[@cattoO_O]

hey Huzefa! I'm practising for my Digital Sat test on October 7 and your videos are really helpful, thanks man ❤

[@ScalarLearning]

You’re very welcome!!!

[@bee_he5741]

We can never find anything better than this ......

[@ScalarLearning]

I'm so glad you found it helpful! Thank you for your support!

[@davidrutitsky9518]

Thank you so so much for these videos. Sometimes I don't understand the khan academy explanations

[@ScalarLearning]

Happy to help!!!

[@umenakid]

Thank you very much mate! You helped a lot!!

[@ScalarLearning]

My pleasure!!

[@oliverocampo2756]

great stuff man!

[@ScalarLearning]

Thank you!

[@pritishrivastav5978]

was looking for y=m(x-a) question on Khan Academy but this too was helpful. Please make a video on that. The question goes as: The equation y= 3/2(x-8) is graphed in the xy plane. Which of the following equations will have a graph that is parallel to the graph of the above equation and have an x-intercept on the negative x-axis? a.) y= 3/2(x+8) b.) y= 3/2x-8 c.) y= -2/3(x+8) d.) y= -2/3x - 8 There is no error in options so don't think I have not included brackets in 2 options.

[@ScalarLearning]

The answer is (A) because to be parallel it must have the same slope and to find the x-intercept, plug 0 in for y and you get -8. Make sense?

[@jeromeagwu8999]

Yes I was here for the same thing thanks

[@nataliaa1746]

do more SAT videos

[@ScalarLearning]

Absolutely! New ones come out every Wednesday

[@tanzinaanisa8473]

Thank you so much!!

[@ScalarLearning]

You're very welcome!!

[@bilalmuhammad5112]

There are 2,3 graph problems in khan academy relative to this topic which are similar to each other but have totally different concepts than rest of questions. If you look after that, it would be very helpful for me and many other students out there. I have s.s of these questions, should I email you or send you in social chat

[@ScalarLearning]

Please email them to me at [e-mail]

[@aliabdulrahim9196]

Huzefa Kapadia

[@ScalarLearning]

That's correct! Thank you for checking out my channel!

[@ShrayDas]

Thank you so much!