hello! i'll be writing SAT tomorrow (jun 1st). just wanted to say you have been a life saver for me for the past 3 weeks! your workings for math are really efficient to understand the concepts :) please wire some bran waves to all the SAT writers tomorrow!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! thank you so much
Question 21 at 42:00 can be solved by equating the 2 missing surface areas (of the square bases) when the 2 prisms are glued together. It was derived that K = 2x^2 + 360x The 2 missing surface areas are 2x^2 = (2/47)K 47x^2 = K Substituting the value of K, 47x^2 = 2x^2 + 360x 45x = 360 x = 8
@@ScalarLearning dude so much overlap with your video and the test! definitely helped me get a higher score. b^2-4ac came clutch a few times in the second module.
One can also find distance of circle with desmos as long as there are two points. For question 20, set a=(-0.5,9.5) then b=(-0.5,-10.5) then do distance(a,b) then you’ll get 20. Distance is like the diameter so remember that then divide by 2 to get the radius, which is 10.
Quick tip for the first one : you can also set the length of arc PQ(=SR) as x, meaning PS(=QR)=2x, meaning 144pi=6x. Set pi aside (as four answers do) and just divide 144 by 6 to find x(=24). And since we're searching for 2x, you multiply 24 by 2 to get 48 (then tack on pi, you get the idea).
Question 22 at 4:00 can be solved by simply setting the discriminant to zero since the intersection occurs at only one point (one solution). This avoids the calculation of the vertex. f(x) = -2x^2 +9x - c/2 Discriminant: 9(9) - 4(2)(c/2) = 0 81 - 4c = 0 c = 81/4
Question 22 at 20:00 can be resolved without DESMOS. It was derived that f(x) = (2/3)x^2 + 4x + c To find the relationship between k and c, we plug in the value of the vertex (h,k) = (-3,k) f(-3) = k = (2/3)9 + 4(-3) + c k = -6 + c c = k + 6 Thus, c
For Question 22 at 1:05:00, it was derived that -3a = (a+b) In order to evaluate (a+b) given that “a” is an integer greater than one, the expression can be rewritten as a = -(a+b)/3 For “a” to be greater than one, (a+b) must be less than -3 In addition, for “a” to be an integer, (a+b) must be a multiple of 3
For Question 22 at 47:00, it was derived that 64a - 14 = (a+b+c) In order to evaluate (a+b+c) knowing that “a” must be greater than zero, the expression can be rewritten as 64a = (a+b+c) - (-14) For “a” to be greater than zero, (a+b+c) must be greater than -14.
Question 20 at 10:30 can be solved QUICKLY by just completing the square. It was derived that 2x^2 - 24x + 64 - a All we have to do is complete the square: 2x^2 - 24x + 64 - a = 2 [ (x - 6)^2 ] Thus, we have x = 6 as the answer. BOOM done!!! If you are interested in the value of “a” 64 - a = 2 (36) a = 64 - 72 = -8 BOOM BOOM done!
At 10:09, an even easier way to use desmos with this problem is to write f(x)=4x^2+64x+262 and in another line, just write g(x)=f(x+5) and it will graph the translated function without having to replace each x :))))
For Question 22 at 30:00, since we are interested in the difference of the mean, we can reduce the number of calculations by directly subtracting integer values that have the same number of frequency in computing the average. ((3(20-19) + 4(30-29) + 7(40-39) + 9(50-49))/23 = (3+4+7+9)/23 = 1 As a sanity check, we can eyeball the graphs and use the process of elimination. Answers A and D are obviously wrong. Both data sets have approximately a mean of 35 (high 30s actual), but with data set A slightly larger than data set B. So answer B is the correct answer.
Question 21 at 53:00 was created by SAT for people to make unnecessary work. They gave the formula, electric flux equals product of electric field and surface area. They also gave the electric field as 29 v/m. But guess what! All you have to do for this problem is to ratio the large area to the total area. 4640 (9x/10x) = 4176 BOOM BOOM BOOM
For 49:36 I think its easier to realize those are similar triangles (2 shared angles) and that they have a scale factor of 7. So with that, AD = 7 root 34 + root 34 which is just 8 root 34 and AE is 24. Now do (8 root 34)^2 - (24)^2 to find that DE^2=1600 and DE=40. Now just do 40*24*.5 and you got 480 boom done.
You can use DESMOS to find the answer, but you have to vary the constant “a” in order to find it. I think you are better off finding ways to simplify your calculations, and then use DESMOS to verify your answer if you still have time. This is one way to simplify your calculations: 2x^2 - 24x + (64-a) = 0 Determinant D: Let m = 64-a D = (24)(24) - 4(2)m = 3(24) - m = 0 From which we have m = 72 You can calculate “a” from “m” if required. In this case, it is not. You can go ahead and solve for x. 2x^2 - 24x + 72 = 0 X^2 - 12x + 36 = 0 (x-6)(x-6) = 0 BOOM! You are done.
Hey for question 22 in practise 1 module 2 could you further explain why option b is incorrect since wouldn’t it also be sufficient. To prove side angle suds
the "sas" values have to be in order - so the angle would have to be sandwiched in between the two corresponding side values. since the angle of 60 isn't in between the corresponding sides, similarity cannot be proven. if it were stated that sides ML and SR were corresponding and similar sides, then similarity could be proven because the 60 degree angle would be in between the corresponding sides.
You are correct if both triangles LMN and RST are right triangles, but they are not. If it is given that angles N and T are equal or angles M and S are equal, then you are also correct because MN and ST are corresponding sides. Answer D is correct. If angle M is 70 and angle T is 50, it means that angle S becomes 180 - 60 - 50 = 70. Angle M and angle S now become corresponding angles. The question is asking for a sufficient condition to show similarity. So you have to look for 2 corresponding angles AA, 3 corresponding sides SSS, or 2 corresponding sides with a corresponding angle in between SAS.
For number 18, could you just plug in x+4 for x and realize that option C is the only possible one since B is a constant? I didn't go through the whole translation thingy but I'm not sure if my way is a correct or I just got it based off of luck.
If you plug in x+4 for x, you have f(x+4) = a / (x+4+b) b is a constant that you have to derive from the graph of f(x). If we pick the following points for f(x) (x,y) = (-6,-3), (-5,-6) We have for a = y (x+b) a = 18 -3b and a = 30 - 6b From which we have a = 6 and b = 4 Hence, f(x) = 6/(x+4) and f(x+4) = 6/(x+8) To summarize, if you pick option C as your answer, you got LUCKY because “b” happens to be equal to 4. If b = -4, option A is the correct answer. And if b =0, option B is the correct answer.
hello! i'll be writing SAT tomorrow (jun 1st). just wanted to say you have been a life saver for me for the past 3 weeks! your workings for math are really efficient to understand the concepts :) please wire some bran waves to all the SAT writers tomorrow!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! thank you so much
brain* oops
Thank you so much for the kind words, best of luck tomorrow!!
what is your score ?
Question 21 at 42:00 can be solved by equating the 2 missing surface areas (of the square bases) when the 2 prisms are glued together. It was derived that
K = 2x^2 + 360x
The 2 missing surface areas are
2x^2 = (2/47)K
47x^2 = K
Substituting the value of K,
47x^2 = 2x^2 + 360x
45x = 360
x = 8
praying for a 1400+ tmrw, thanks for the tips very helpful!
Love that as a goal, good luck!!
@@ScalarLearning dude so much overlap with your video and the test! definitely helped me get a higher score. b^2-4ac came clutch a few times in the second module.
1440 😊
@@soccerboigronk11 nice job bro, aiming for above a 1400 aswell, any tips you have for me?
@@soccerboigronk11 how do you recommend practicing for the math section?
thank you !! aiming for a 1530+ from a 1480 tmr!
You got this!
same good luck broski
what did u get
@@elflonghand5056 1550 w an 800 in math! this guy rlly helped
you will definitely get it, best of luck
One can also find distance of circle with desmos as long as there are two points. For question 20, set a=(-0.5,9.5) then b=(-0.5,-10.5) then do distance(a,b) then you’ll get 20. Distance is like the diameter so remember that then divide by 2 to get the radius, which is 10.
Quick tip for the first one : you can also set the length of arc PQ(=SR) as x, meaning PS(=QR)=2x, meaning 144pi=6x. Set pi aside (as four answers do) and just divide 144 by 6 to find x(=24). And since we're searching for 2x, you multiply 24 by 2 to get 48 (then tack on pi, you get the idea).
yep, i did the same
Question 22 at 4:00 can be solved by simply setting the discriminant to zero since the intersection occurs at only one point (one solution). This avoids the calculation of the vertex.
f(x) = -2x^2 +9x - c/2
Discriminant:
9(9) - 4(2)(c/2) = 0
81 - 4c = 0
c = 81/4
Question 22 at 20:00 can be resolved without DESMOS. It was derived that
f(x) = (2/3)x^2 + 4x + c
To find the relationship between k and c, we plug in the value of the vertex (h,k) = (-3,k)
f(-3) = k = (2/3)9 + 4(-3) + c
k = -6 + c
c = k + 6
Thus, c
1st problem. Arc PS = 2x b/c arc PQ = x. Arc RQ = 2x, too. Arc SR = x. Circum = 6x = 144pi. Solve for x (24pi). Double for arc RQ. 48pi.
Legend
Thank you for this. I thought I was crazy when I saw his more complicated explanation.
For Question 22 at 1:05:00, it was derived that
-3a = (a+b)
In order to evaluate (a+b) given that “a” is an integer greater than one, the expression can be rewritten as
a = -(a+b)/3
For “a” to be greater than one, (a+b) must be less than -3
In addition, for “a” to be an integer, (a+b) must be a multiple of 3
For Question 22 at 47:00, it was derived that
64a - 14 = (a+b+c)
In order to evaluate (a+b+c) knowing that “a” must be greater than zero, the expression can be rewritten as
64a = (a+b+c) - (-14)
For “a” to be greater than zero, (a+b+c) must be greater than -14.
Question 20 at 10:30 can be solved QUICKLY by just completing the square. It was derived that
2x^2 - 24x + 64 - a
All we have to do is complete the square:
2x^2 - 24x + 64 - a = 2 [ (x - 6)^2 ]
Thus, we have x = 6 as the answer.
BOOM done!!!
If you are interested in the value of “a”
64 - a = 2 (36)
a = 64 - 72 = -8
BOOM BOOM done!
At 10:09, an even easier way to use desmos with this problem is to write f(x)=4x^2+64x+262 and in another line, just write g(x)=f(x+5) and it will graph the translated function without having to replace each x :))))
For Question 22 at 30:00, since we are interested in the difference of the mean, we can reduce the number of calculations by directly subtracting integer values that have the same number of frequency in computing the average.
((3(20-19) + 4(30-29) + 7(40-39) + 9(50-49))/23
= (3+4+7+9)/23 = 1
As a sanity check, we can eyeball the graphs and use the process of elimination. Answers A and D are obviously wrong. Both data sets have approximately a mean of 35 (high 30s actual), but with data set A slightly larger than data set B. So answer B is the correct answer.
Thank you sm!! I'm taking it tomorrow, hoping to get a 1400+. Also, off topic but your eyes are very pretty
Question 21 at 53:00 was created by SAT for people to make unnecessary work. They gave the formula, electric flux equals product of electric field and surface area. They also gave the electric field as 29 v/m. But guess what! All you have to do for this problem is to ratio the large area to the total area.
4640 (9x/10x) = 4176
BOOM BOOM BOOM
For 49:36 I think its easier to realize those are similar triangles (2 shared angles) and that they have a scale factor of 7. So with that, AD = 7 root 34 + root 34 which is just 8 root 34 and AE is 24. Now do (8 root 34)^2 - (24)^2 to find that DE^2=1600 and DE=40. Now just do 40*24*.5 and you got 480 boom done.
Legend!!
ur a lifesaver
I had few questions similar to these in my june SAT. Math was easy
Legend!!!
these videos are really helpful. thank you!
You’re very welcome!!
Watching your videos, i could get a 790 on math. Thank you very much!!!
That's great!
You are a life saver
Glad you found it useful!!
thank you, praying for a 1350 tomorrow
Good luck Davida!!
I’m praying for a 900+
What'd you get?
@@estebancasillas9529 i got a 1370! 710 on the math section
thank you!
You’re welcome!!
Good Day. But the 1st question's answer should be 96pie instead of 48. As QR is double. Please have a look into this. Thanks
Check again, bc Qr is double PQ, and PQ is only 24pi. Take another look
17:46 why can’t the answer be 0 since the factors of the equation are (2)(x+0)(x+13)(x+8)
0 is not positive
great vid
@12:42, is there a way we can get the answer by plugging into desmos?
Yes most likely, about to drop a Desmos video today
You can use DESMOS to find the answer, but you have to vary the constant “a” in order to find it. I think you are better off finding ways to simplify your calculations, and then use DESMOS to verify your answer if you still have time. This is one way to simplify your calculations:
2x^2 - 24x + (64-a) = 0
Determinant D:
Let m = 64-a
D = (24)(24) - 4(2)m = 3(24) - m = 0
From which we have m = 72
You can calculate “a” from “m” if required. In this case, it is not. You can go ahead and solve for x.
2x^2 - 24x + 72 = 0
X^2 - 12x + 36 = 0
(x-6)(x-6) = 0
BOOM! You are done.
yes you can use desmos.
Hey for question 22 in practise 1 module 2 could you further explain why option b is incorrect since wouldn’t it also be sufficient. To prove side angle suds
LEGEND ❤.
9:17 wouldn't the answer be B??
RT= 30
LN= 10
By this way
If MN = 7
ST would be 7(3)= 21?? I'm a bit confused
the "sas" values have to be in order - so the angle would have to be sandwiched in between the two corresponding side values. since the angle of 60 isn't in between the corresponding sides, similarity cannot be proven. if it were stated that sides ML and SR were corresponding and similar sides, then similarity could be proven because the 60 degree angle would be in between the corresponding sides.
You are correct if both triangles LMN and RST are right triangles, but they are not. If it is given that angles N and T are equal or angles M and S are equal, then you are also correct because MN and ST are corresponding sides.
Answer D is correct. If angle M is 70 and angle T is 50, it means that angle S becomes 180 - 60 - 50 = 70. Angle M and angle S now become corresponding angles.
The question is asking for a sufficient condition to show similarity. So you have to look for 2 corresponding angles AA, 3 corresponding sides SSS, or 2 corresponding sides with a corresponding angle in between SAS.
For number 18, could you just plug in x+4 for x and realize that option C is the only possible one since B is a constant? I didn't go through the whole translation thingy but I'm not sure if my way is a correct or I just got it based off of luck.
If you plug in x+4 for x, you have
f(x+4) = a / (x+4+b)
b is a constant that you have to derive from the graph of f(x).
If we pick the following points for f(x)
(x,y) = (-6,-3), (-5,-6)
We have for a = y (x+b)
a = 18 -3b and a = 30 - 6b
From which we have
a = 6 and b = 4
Hence,
f(x) = 6/(x+4) and
f(x+4) = 6/(x+8)
To summarize, if you pick option C as your answer, you got LUCKY because “b” happens to be equal to 4. If b = -4, option A is the correct answer. And if b =0, option B is the correct answer.
Legend
Thank you!!
Will I get my result right away tommorow?
You mentioned a easy test means less of a curve so does that mean we will have to wait for the curve to be released? Thanks!
It comes out after 2 weeks
Usually around 13 days after you take it
always with the big ass hats
46:00, 47:30 s