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Dear Julia ,
If you are join XYZ college and complete your 1st year course but if you will take admission in another college without informing XYZ college then it will trouble in future because you have already student of this college or university with particular registration number and it is one type of illegal work .So you should confirm to college and leave with the formalities then only you will not facing any trouble in future .
Hope it will helpful
Thanks
Hey Jaywadmare, the anwer of your question is 34/3.
Also go through following explanation.
Explanation : a^2 = d ^12
b^3 = d ^12
c^9 = d ^12
after that logd ( d^12) 1/2 + 1/3 + 1/9 = logd (d^12)^ 51/54
= logd (d ^12 * 17/18)
= 1 * 12 * 17/18
= 34/3.
I hope this will helps to you and thank you !!!
In a triangle ABC cos(A+B)= cosAcosB-sinAsinB. cos(A-B)= cosAcosB+sinAsinB.
there by, by adding these two and multiplying the result by 2 times you get the desired answer. 4cosAcosB = 2[cos(A+B) + cos (A-B)]. This can also be done by interchanging the signs and the answer will remain the same.
Hey!
NO, I really dont think modern ABC is the appropriate book for NEET prep… Though it can help you with the 12 grade exams.
For NEET-UG,
1st, For the KNOWLEDGE part -primarily you should go for NCERT first surely and after that for extra knowledge Allen packages or Trueman’s for biology would be more than sufficient.
2nd, For the APPLICATION part- buy sample paper books of previous years question papers to practice ,after retaining the theoretical content.
Apply ur knowledge in the questions…!!
Using distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) to find distance between two points A(x1, x2) and B(y1, y2), we can find the distance between A and B, B and C and A and C respectively.
We have, AB = sqrt((3-1)^2 + (5-k)^2)
= sqrt(2^2 + (5-k)^2
= sqrt(4 + (5-k)^2) (Equation 1)
BC = sqrt((6-3)^2 + (2-5)^2)
= sqrt(3^2 + (-3)^2)
= sqrt (18) (Equation 2)
AC = sqrt((6-1)^2 + (2-k)^2)
= sqrt(5^2 + (2-k)^2)
= sqrt((25 + (2-k)^2) (Equation 2)
Using Pythagoras theorem, we have AC^2 = AB^2 + BC^2
Substituting the values of AB, BC and AC from equation 1, 2 and 3 respectively, we have:
(sqrt((25 + (2-k)^2)) ^2 = (sqrt(4 + (5-k)^2)) ^2 + (sqrt (18))^2
= 25 + (2-k)^2 = 4 + (5-k)^2 + 18
= 25 + 4 + k^2 – 4k = 4 + 25 + k^2 -10k + 18
Rearranging and solving, we get, K = 3 (Answer)
Greetings Aspirant,
If the magnetic field along AB is B , the force acting on BC=F=BiL
Now the length of the side AC is Root"2L"
Also the current reverses the direction in this section as compared to that in BC.
Hence force on AC= Root"2"L( B× I )
= − Root"2"BILsin(45)
=−BIl=− F
If you are unable to understand the Answer I have uploaded an Image for Better Understanding.
https://drive.google.com/file/d/1e6gsucWKEXxS1si-XlTXIEmc11SRBq8K/view?usp=sharing
Hope My Answer Helped You.
Best Of Luck For Future.
Since A, B , C are the angles of a triangle and we know that the sum of all angles in a triangle is 180 degrees .
Consider A =5x
B=1x
C=6x
5x+1x+6x=180
12x=180
x=15 degrees
Angles A=5 X 15=75 degrees
B=15 degrees & C= 6X15= 90 degrees
By sine rule we know that (a/sinA)=(b/sinB)=(c/sinc)=2R
hence a=2RsinA, b=2RsinB, c=2RsinC
a:b:c=2RsinA : 2RsinB : 2RsinC
=sinA : sinB : sinC
=sin75 : sin15 : sin90
Hi
As per your question the solutions is:
It is for the reason that ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.∴ AB/PQ = AC/PR = BC/QR
...(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R …(ii)
Since AD and PM are medians, they may divide their contrary sides.∴ BD = BC/2 and QM = QR/2 ...(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ...(iv)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (ii)]
AB/PQ = BD/QM [Using equation (iv)]
∴ ΔABD ~ ΔPQM (By SAS similarity criterion)⇒ AB/PQ = BD/QM = AD/P
Hope helpful.
Hey there
You have good marks in your 10th and 12th but little low in your graduation so you need to get very good percentile in CAT for general category. CAT is all about speed so you have time in this lockdown you can speed up all the three sections specially quantitative analysis.
Check CAT cutoff in details
https://bschool.careers360.com/articles/cat-cutoff/amp
Good luck
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