for the second question, i used c/a as the formula for the product of the solutions of a quadratic equation, ab/57 = kab, everything cancels and simplifies real nicely and you end up with k = 1/57
For the first question 4:30, b has to be less than 5. Only b = -6.7 is the valid solution. The other b value is extraneous. 5 = a^n + b The min value of a^n is zero, which gives the max value of b as 5. In the calculation of b, a^n disappeared from the equation with the substitution a^n = 5 - b, resulting in the extraneous value of b.
2nd question you can do it by using product of solution=c/a (constant/coefficent of x^2). consequently,you'll get kab=ab/57 ... cancel out ab from both side you'll get k=1/57 this is a much easier appproach
The first question can be easily solved by desmos by first creating a table and in place of the eqn, we should write y1~a^x1+b, the we can easilyyy keep the value of n as 1 or 2 or 3 or 4. the value of b remains unchanged.. HOPE THIS HELPS
Thanks for the video. In the third question, the underlying concept is that Sin(180 - theta) = sin(theta). That is why, the Sine of the reference angle that you mentioned in your solution is equal to sq.root 3 divided by two --> the given value of Sine for the given angle (theta) in the question.
Hard is a relative term. 1) b < 5, so only the negative possibility makes sense. Note: had the difference in y-values been the product of consecutive integers, the quadratic would factor ; 2) trivial by Vieta's formulas; 3) trivial from unit circle trig values.
For the 1st question: What if we follow this approach: we substitute the given points and add the two equations while multiplying the first equation by -1. We'll have something like this: (a^{2n} - a^n - 125 = 0), and then we assume that (a^n = x). Solving for (x), we get (x^2 - x - 125 = 0). Then (a^n = x) will be one of these two values: 11.692 or -10.692. If we take the first value and substitute it into the equation:(5 = a^n + b), we find that (b = -6.692).
there's 2 ways of using simultaneous equations, either elimination OR substitution, she used subtitution but you use elimination, its best practice to use substitution esp considering theres a quadratic involved, I see what you did and that's what I initially did, but I felt as thought it would look messy so I did subtitution, however your method should work tho
@@Faizanshaikh-i7h CAST diagram, cosine is negative in the 2nd quadrant, so if cos(60) = 1/2, cos(120) = -1/2 as 120 is in the 2nd quadrant where only sine is positive
How does rounding works?, for example, for the first example that you showed, can I just write 6.69? Or how many 'significant figures' must the answer have? (if it's not detailed by the question)
For question 1: Why can't I divide the second equation 130=a^2n + b by the first equation 5=a^n + b? Wouldn't this give me 26 = a^n + 1 and 25 = a^n which allows me to plug back in and solve for b?
@@captainbizarre3523 Theta is more than 90 degrees but less than 180 degrees. That is why it is in the second quadrant. The reference angle is 180 degrees minus theta, and cosine is negative in the second quadrant.
@@sedramohammed1264 Sure, if you create a table in Desmos using the points given in the problem (n,5), (2n,130) and create your own value for n (can be whatever). Then type Y1~a^x1+b it should just give you the value b= -6.69151
@@sedramohammed1264First, make a table. Input n and 2n in the x column, input 5 and 130 in the y column, it will show an error because n is not defined, add a line saying n = 1, this will create a slider for n go to the next line and write this equation y1~(a^x1)+b, make sure to use ~ and not =, also make sure to put y1 and x1, not y and x this is so that your values correspond to the ones in the table when you do this you will get the regression values, regardless of n, your b value remains the same, -6.692
so in question 1 i found the slope witch was 125/n then i removed n [(2n-n) and (5-125)] from (n,5) ------> (0 , -120 ) so b = -120 why is my answer wrong
Are these seriously considered hard problems?? The solutions are pretty straightforward and I can recall that we were solving problems like that as common homework, when I was in high school!
For the third question, your diagraming is technically wrong because the radius of the circle is 1 and your diagonal was 2. A better way to teach it is using the unit circle.
yeah if r=1 then how does one side of the triangle where its on the radius equal to 1 even though its not the length of the radius olso how is the angle next to Q olso Q they don't look equal
Theta is drawn in the second quadrant. The reference angle in the second quadrant is 180 - theta, referenced to the x axis. Sin (theta) = sin (180-theta).
search up "A level further maths past papers" and do those questions if you want "hard", alternatively search up "MAT past papers" which are similar in style to the SAT
for the second question, i used c/a as the formula for the product of the solutions of a quadratic equation, ab/57 = kab, everything cancels and simplifies real nicely and you end up with k = 1/57
Love this!! ❤️
This was really helpful. Good job.
👍helpful
Wonderful scholar!
I tought scalar learning this.
For the first question 4:30, b has to be less than 5. Only b = -6.7 is the valid solution. The other b value is extraneous.
5 = a^n + b
The min value of a^n is zero, which gives the max value of b as 5. In the calculation of b, a^n disappeared from the equation with the substitution a^n = 5 - b, resulting in the extraneous value of b.
2nd question you can do it by using product of solution=c/a (constant/coefficent of x^2).
consequently,you'll get kab=ab/57 ... cancel out ab from both side you'll get k=1/57
this is a much easier appproach
The first question can be easily solved by desmos by first creating a table and in place of the eqn, we should write y1~a^x1+b, the we can easilyyy keep the value of n as 1 or 2 or 3 or 4. the value of b remains unchanged..
HOPE THIS HELPS
yep it helped
Thanks for the video. In the third question, the underlying concept is that Sin(180 - theta) = sin(theta). That is why, the Sine of the reference angle that you mentioned in your solution is equal to sq.root 3 divided by two --> the given value of Sine for the given angle (theta) in the question.
for the 3rd question isn't it possible to just use inverse functions to find the angle then just finidng the cos of the angle?
Hard is a relative term. 1) b < 5, so only the negative possibility makes sense. Note: had the difference in y-values been the product of consecutive integers, the quadratic would factor ; 2) trivial by Vieta's formulas; 3) trivial from unit circle trig values.
in the first question cant we replace n by any number and it on desmos using a regression?
Hey ,
Can you make a video on June Expected questions and Predictions ?
i also want that
Hey how do I plug the desmos equation in for the second question. Its not working for me
For the 1st question: What if we follow this approach: we substitute the given points and add the two equations while multiplying the first equation by -1. We'll have something like this: (a^{2n} - a^n - 125 = 0), and then we assume that (a^n = x). Solving for (x), we get (x^2 - x - 125 = 0). Then (a^n = x) will be one of these two values: 11.692 or -10.692. If we take the first value and substitute it into the equation:(5 = a^n + b), we find that (b = -6.692).
there's 2 ways of using simultaneous equations, either elimination OR substitution, she used subtitution but you use elimination, its best practice to use substitution esp considering theres a quadratic involved, I see what you did and that's what I initially did, but I felt as thought it would look messy so I did subtitution, however your method should work tho
For second question use vietas formula
Thanks for posting the "unit circle" question. My students have been saying that's been appearing, but I wasn't sure exactly how.
In the third question, wasn’t the angle given in the first quarter?
No,the angle is greater than 90 therefore the theta lies in 2nd quadrant
@@Faizanshaikh-i7h CAST diagram, cosine is negative in the 2nd quadrant, so if cos(60) = 1/2, cos(120) = -1/2 as 120 is in the 2nd quadrant where only sine is positive
How does rounding works?, for example, for the first example that you showed, can I just write 6.69? Or how many 'significant figures' must the answer have? (if it's not detailed by the question)
Sorry, it should be -6.69
This isn't a realistic question. The test would always tell you the exact decimal value to round to.
For question 1: Why can't I divide the second equation 130=a^2n + b by the first equation 5=a^n + b? Wouldn't this give me 26 = a^n + 1 and 25 = a^n which allows me to plug back in and solve for b?
i did that and apparently and I got different values of B for each equation
Your division is not correct,
(a^n + 1)(a^n + b) = a^2n + (b+1)a^n + b
For the last question, how do you know the reference angle is sqrt3/2? thats the angle of theta.. isnt the reference angle different??
The reference angle in the second quadrant is 180 - theta. Sin (theta) = sin (180 - theta) = rad(3)/2
Why was question 3 in quadrant 2? how we know which quadrant to work in?
The question is asking:
In the figure, theta is an angle.
The figure is showing second quadrant. So you work in the second quadrant.
@@OverclockingCowboyi’m confused how you know to work in q2, since the theta in the equation looks like it’s for the large angle
@@captainbizarre3523
Theta is more than 90 degrees but less than 180 degrees. That is why it is in the second quadrant.
The reference angle is 180 degrees minus theta, and cosine is negative in the second quadrant.
Btw for the first one you can just use a table on Desmos and do exponential regression 👍
Can u explain a bit
@@sedramohammed1264 Sure, if you create a table in Desmos using the points given in the problem (n,5), (2n,130) and create your own value for n (can be whatever). Then type Y1~a^x1+b it should just give you the value b= -6.69151
@@sedramohammed1264First, make a table. Input n and 2n in the x column, input 5 and 130 in the y column,
it will show an error because n is not defined, add a line saying n = 1, this will create a slider for n
go to the next line and write this equation y1~(a^x1)+b, make sure to use ~ and not =, also make sure to put y1 and x1, not y and x
this is so that your values correspond to the ones in the table
when you do this you will get the regression values, regardless of n, your b value remains the same, -6.692
so in question 1 i found the slope witch was 125/n
then i removed n [(2n-n) and (5-125)] from (n,5) ------> (0 , -120 ) so b = -120
why is my answer wrong
Are these seriously considered hard problems?? The solutions are pretty straightforward and I can recall that we were solving problems like that as common homework, when I was in high school!
great video， i got a awesome score for that
The second question,
c/a = product of zeroes
For the third question, your diagraming is technically wrong because the radius of the circle is 1 and your diagonal was 2. A better way to teach it is using the unit circle.
She is not using the unit circle. She is using the given value of sine theta which is equal to rad(3)/2. It is more convenient.
yeah if r=1 then how does one side of the triangle where its on the radius equal to 1 even though its not the length of the radius olso how is the angle next to Q olso Q they don't look equal
Is there any chance to get derivatives on Math?
Absolutely not
sat can for university it tenology
When will your june course drop
Predictions coming Wednesday!
i don't get what you did on question 3, wasn't teta already drawn? why did you assume teta was other angle
Theta is drawn in the second quadrant. The reference angle in the second quadrant is 180 - theta, referenced to the x axis.
Sin (theta) = sin (180-theta).
Why is preptly not free....
how are these hard??
search up "A level further maths past papers" and do those questions if you want "hard", alternatively search up "MAT past papers" which are similar in style to the SAT
U really call this hard even boards maths is better than this