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SAT Exam June 2024 Predictions
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- Опубликовано: 20 май 2024
- June 2024 Digital SAT Predictions
In this video, Jackie jumps into her predictions for the June 2024 Digital SAT Exam.
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On the greatest possible value of b question 9:30, we can first examine the special case (easier) and then the general case (harder).
Special Case
(34x + m) (x + n)
where mn = 70
We have b = 34n + m
MAX b occurs at m=1 and n=70
MIN b occurs not at m=70 and n=1
but at m=35 and n=2
General Case
(px + m) (qx + n)
It is advisable to stop here. You will run out of time.
pq = 34 and mn = 70
We have b = pn + mq
b = (34/q)(70/m)+ mq
Max b can easily be found as
q = m = 1 and
p = 34, n = 70
Min b is harder to find.
We can rearrange b as
b = (34/q) n + (70/n) q
We can minimize with n/q = 2
Resulting in
p=34, q=1, m=35, n=2
[requires additional work]
19:24
On the (x-9) factorization 15:30, we may apply the Zero Remainder Theorem since the factors are given.
Case I. (x - 9)
Not a factor. There is no need to test as the remainder is obviously large for x = 9.
4(81) + 31(9) - 45
Case II. (4x - 5)
4(5/4)(5/4) + 31(5/4) - 45
25/4 + 155/4 - 45
45 - 45
It is a factor
Long division for case II
1. 9
4 - 5 // 4 + 31 - 45
4 - 5
-------
36 - 45
The factors are
(4x - 5) (x + 9)
Another great way!
On the (x + 2b) factor 19:40, since you already made the other factor as (3x + k), you might as well make it as (3x + 9). All the answer choices have 18b as the last term. Only the middle term is changing.
(3x + 9) (x + 2b)
The middle term after expanding is
6b + 9
Using different values of b as a positive integer, the middle term becomes
15, 21, 27, 33, 39, 45
Thus, Answer Choice D.
This is similar to what you have done. But it may speed things up.
Great channel and you actually cover some of the hardest questions unlike some of the shorts I’ve seen.
Thank you so much! =)
THANK YOU.Your teching methods are realllly good
Thank you! I hope it helps if you are taking the Digital SAT =)
Thank you.😊 the video was so informative!
You're welcome! Glad it was informative =)
Thank you for this prediction and continue making helpful videos.
You're welcome! Hope it helps!! =)
Thank you so much for your time, so grateful
Of course! I am here to help. Hope the video helps you =)
in the minute 27:00 when you said which is you were very cute 😂😍😍thank you for the effort
haha! I do it subconsciously😅
very useful thank u so much
You're welcome! Glad you found it useful! =)
Could you explain why c = 70 in problem 3. I didn't really understand why c had to be equal to 70 for b to be a positive integer
C doesn't have to be 70 to be a positive integer...it has to be 70 to be the GREATEST positive integer...I will try to explain more here =) :
So we expand: (gz^7 + r)(qz^7 + c) = gqz^14 + gcz^7 + rqz^7 + rc
Coefficient of z^14: gq = 34
Coefficient of z^7: gc + rq = b
Constant term: rc = 70
Find g and q:
Since g and q are positive integers such that gq = 34, the possible pairs (g, q) are:
(g, q) = (1, 34), (2, 17), (17, 2), (34, 1)
Find r and c:
Since rc = 70, the pairs (r, c) can be:
(1, 70), (2, 35), (5, 14), (7, 10), (10, 7), (14, 5), (35, 2), (70, 1)
We need to find which combination of these pairs will give the MAXIMUM value of b.
Calculations for Each Pair (g, q):
Here I can test each combination of (g, q) with (r, c):
For (g, q) = (1, 34):
For (r, c) = (1, 70):
b = gc + rq = 1 * 70 + 34 * 1 = 70 + 34 = 104
For (r, c) = (2, 35):
b = gc + rq = 1 * 35 + 34 * 2 = 35 + 68 = 103
For (r, c) = (5, 14):
b = gc + rq = 1 * 14 + 34 * 5 = 14 + 170 = 184
For (r, c) = (7, 10):
b = gc + rq = 1 * 10 + 34 * 7 = 10 + 238 = 248
For (r, c) = (10, 7):
b = gc + rq = 1 * 7 + 34 * 10 = 7 + 340 = 347
For (r, c) = (14, 5):
b = gc + rq = 1 * 5 + 34 * 14 = 5 + 476 = 481
For (r, c) = (35, 2):
b = gc + rq = 1 * 2 + 34 * 35 = 2 + 1190 = 1192
For (r, c) = (70, 1):
b = gc + rq = 1 * 1 + 34 * 70 = 1 + 2380 = 2381
So, conclusions....we need to find the GREATEST integer value of B...and for that reason c=70 .... I hope this helps to make a bit more sense? =)
Ya why?
Could anyone tell me if these predictions were correct on what was actually on the exam
I watched this video before my SAT in June and can confirm her predictions are good. At least they were for June. She uses very similar questions that have been on past official exams and SAT has been reusing questions from exams. All of these concepts were on June exam. For me, she is the best at prediction but I will see for August. Hope it helps.
Really happy to hear this! Thank you so much! =)
I love when u say "isnt that amazing" lol
hahha! Thank you 😂🥰🥰
Nervous, but excited. Bad case scenario i get a 1400-1440. Middle case scenario I get a 1450-1490. Good case scenario I get 1500+. I got a 1510 on a practice test but I don’t want to get my hopes up just yet.
Good luck! I hope it is 1500+ =)
for the first question, isnt 0 also a solution. so wouldn't 0 be the smallest value of b
0 is an integer...However, it asks for the smallest POSITIVE integer...and 0 is neither negative nor positive. Also, unrelated to this problem but an interesting number property of 0 that 0 is even. Hope this helps! 🤓
@@epicexamprep Thank you so much
for the question at 24:00, would it be easier to plug in all equations into desmos and create a positive integer slider for b from 1-50 (or whatever number) and will be able to see that the only time a function has a solution at a positive integer equal to x+2b is when b is 5 for 3x^2 +39x +18b. or does this only happen to work in this scenario. thank you
Yes! That would also work! There are a couple different methods to solve this question. I just showed the one that I think is the fastest/simplest way to solve. But to explain further on desmos:
Input the Equation:
For example, for Option D (correct answer), type: y = 3x^2 + 39x + 18b.
After typing the equation, simply type b on a new line and Desmos will prompt you to add a slider. Click on "Add Slider."
Plot the Vertical Line..since we are told factor x+2b=0, we know that x=-2b, so we can type that into desmos:
x=−2b
Drag the slider for b and observe the changes in the graph.
Check for intersections where the graph of the quadratic equation meets the x-axis at
x=−2b.
And we see that b=5.
*Using a slider helps you explore a range of values and visually identify when the polynomial intersects the x-axis at the required point, confirming the factorization condition.***
/////
Alternatively, I could have also plugged in -2b for x in each of the answer choices and see which one equates to 0 because for a polynomial to be divisible by x+2b, substituting
x=−2b into the polynomial should result in zero.
Option A: 3x^2 + 9x + 18b
Substitute x = -2b:
3(-2b)^2 + 9(-2b) + 18b
Simplify:
3(4b^2) - 18b + 18b
Result:
12b^2
Set to zero:
12b^2 = 0
Solve:
b = 0
Since b must be a positive integer, b=0 doesn't work.
Option B: 3x^2 + 24x + 18b
Substitute x = -2b:
3(-2b)^2 + 24(-2b) + 18b
Simplify:
3(4b^2) - 48b + 18b
Result:
12b^2 - 30b
Set to zero:
12b^2 - 30b = 0
Factor:
b(12b - 30) = 0
Solve:
b = 0 or 12b - 30 = 0
12b = 30
b = 30/12 = 2.5
Since b must be a positive integer and this is a non-integer it doesn't work.
Option C: 3x^2 + 30x + 18b
Substitute x = -2b:
3(-2b)^2 + 30(-2b) + 18b
Simplify:
3(4b^2) - 60b + 18b
Result:
12b^2 - 42b
Set to zero:
12b^2 - 42b = 0
Factor:
b(12b - 42) = 0
Solve:
b = 0 or 12b - 42 = 0
12b = 42
b = 42/12 = 3.5
Since b must be a positive integer, this one doesn't work.
Option D: 3x^2 + 39x + 18b
Substitute x = -2b:
3(-2b)^2 + 39(-2b) + 18b
Simplify:
3(4b^2) - 78b + 18b
Result:
12b^2 - 60b
Set to zero:
12b^2 - 60b = 0
Factor:
b(12b - 60) = 0
Solve:
b = 0 or 12b - 60 = 0
12b = 60
b = 5
Since b = 5 is a positive integer solution, Option D confirms that x + 2b is a factor.
I hope this helps! Thanks for your question. =) =)
@owen your method is much faster I just tried it on desmos.
but both ways work fine so rly just pick what you're comfortable with
For the 2nd factoring question, I got 3x(x+27)(x-7) before watching you do it. Is this still correct and the same as 3x-7? (from answers given, I would have chosen B anyways) but I want to know if I technically got the right answer even though yours was different?
Hi! That would not be correct because,
First you can expand (x + 27)(x - 7):
(x + 27)(x - 7) = x^2 - 7x + 27x - 189 = x^2 + 20x - 189
Then, multiply everything by 3x:
3x(x^2 + 20x - 189) = 3x^3 + 60x^2 - 567x
The polynomial 3x^3 + 60x^2 - 567x is cubic, which is different from the original quadratic polynomial 3x^2 + 20x - 63. ...
For these ones for the "why" behind it, use the steps I outline in the video. BUT if you are in a hurry on the exam, definitely plug it into desmos and check out the x-intercepts! I hope this helps!! =) =)
helloo, thanks for the great video but in the third question, can I make z^7=x and then factor them in a way to b become the max.34x^2 would be 34x and x and 70 would be 70 to 1 so 70*34+1*1 would be b.
Hello! Sorry for my delay!! I did not see this comment! Okay...so if I am understanding you correctly, yes that way could also make sense:
34z^14 + bz^7 + 70
So, you are saying z^7 = x. This transforms the polynomial into:
34x^2 + bx + 70
Then we must factor this polynomial such that b is maximized. So we can express the polynomial as a product of two factors:
34x^2 + bx + 70 = (Ax + B)(Cx + D)
Expanding the right side, we get:
ACx^2 + (AD + BC)x + BD
Now we match the coefficients with the original polynomial:
The coefficient of x^2 is AC:
AC = 34
The coefficient of x is AD + BC:
AD + BC = b
The constant term is BD:
BD = 70
To maximize b, we need to find suitable values of A, C, B, and D that satisfy these conditions.
Consider possible factor pairs for 34 and 70:
For 34:
(A, C) = (1, 34), (2, 17), (17, 2), (34, 1)
For 70:
(B, D) = (1, 70), (2, 35), (5, 14), (7, 10), (10, 7), (14, 5), (35, 2), (70, 1)
To maximize b, we need the combination that produces the highest value.
We take A = 34, C = 1 and B = 1, D = 70:
AC = 34 * 1 = 34
BD = 1 * 70 = 70
Calculate b:
b = AD + BC
b = 34 * 70 + 1 * 1
b = 2380 + 1
b = 2381
I think this is what you were asking, right?? Let me know! Hope this made sense!!! =)
@@epicexamprep yeah exactly, thank you for responsee 🥰
11:20 how did you get this lol?? its a may SAT question
😉
SAME!!!!
Hi Mam, is the gap year student eligible to take the SAT exam.
Hello! Yes, a gap year student is eligible to take the SAT exam. There are no specific restrictions on who can take the SAT based on their educational status or age. As long as you meet the registration requirements and pay the necessary fees, you can take the SAT. This applies to high school students, gap year students, and even individuals who have already graduated from high school...hope this answers your question! =)
@@epicexamprep Thank you so much, mam please give some free resources for the SAT exam.
@@Tariqshah-ql6sh Watch all of our videos! They are free! And be sure to do the BlueBook Exams (all 6) as well as the question bank offered on collegeboard.org