(y6) (y5) = y11
Without any equality signs and not knowing the plus or minus values of the given terms they can't be considered to be straight line equations and so therefore no solutions are possible.
y11
If you mean 6+y = 11 then y = 5
[object Object]
(y6) (y5) = y11
49 = 16 + 3ysubtract 16 from each side49 - 16 = 16 - 16 + 3y33 = 3ydivide both sides integers by 333/3 = (3/3)y11 = y==========checks
Without any equality signs and not knowing the plus or minus values of the given terms they can't be considered to be straight line equations and so therefore no solutions are possible.
Dextron III ATF Fluid
hardenhuish is a school with 6 different years, y7, y8, y9, y10, y11 and the 6 forms y12 and y13.
(x + y)(x10 - x9y + x8y2 - x7y3+ x6y4 - x5y5 + x4y6 - x3y7 + x2y8 - xy9 + y10)
me:) nah its just a joke sorry i dont know these im in yr y11 4 gods sake plz forgive me:)
Looking for the same thing! Unit one exams in January ehh?!? Year 10. I'm not I'm in y11 :L
y-parameters :- I1=Y11V1 + Y12V2 I2=Y21V1 + Y22V2 so to obtain Y11 and Y21 make V2=0 , (shorting the output terminals) Y11=I1/V1 Y21=I2/V1 so to obtain Y12 and Y22 make V1=0 , (shorting the input terminals) Y12=I1/V2 Y22=I2/V2 as all the admittance parameters obtained are due to either shorting the input or shorting the output terminals so we ca;; these parameters as short circuit admittance parameters