the same question……
x + 7y = 33 (Eqn 1) 3x + 2y = 7 (Eqn 2) Multiply Eqn 1 by 3, now the equations are 3x + 21y = 99 (Eqn 3) 3x + 2y = 7 (Eqn 2) Subtract Eqn 2 from Eqn 4 and the result of subtraction is 19y = 92 Hence y = 92/19 Now substitute the value of y in Eqn 1 x + (7) x (92/19) = 33 x = 33 - 644/19 = -(17/19)
Eqn (A): => 2x + 5y = 16 Eqn (B): => 5x + 2y = -2 5*Eqn (A) - 2*Eqn (B): 21y = 84 => y = 4 Substituting for y in Eqn (a): x = -2
Eqn 1: x - 2y = 3 Eqn 2: 5x + 3y = 2 5 times Eqn 1: 5x - 10y = 15 Eqn 2 - 5*Eqn 1: 0x + 13y = -13 so that 13y = -13 or y = -1 Substitute in Eqn 1: x + 2*1 = 3 x + 2 = 3 x = 1 Solution: x = 1, y = -1
c - 5d = 2 2c + d = 4 Multiply eqn 1 by 2: 2c - 10d = 4 Subtract this from eqn 2: 0c + 11d = 0 which implies that d = 0 Then, by eqn 1, c = 2
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x + 7y = 33 (Eqn 1) 3x + 2y = 7 (Eqn 2) Multiply Eqn 1 by 3, now the equations are 3x + 21y = 99 (Eqn 3) 3x + 2y = 7 (Eqn 2) Subtract Eqn 2 from Eqn 4 and the result of subtraction is 19y = 92 Hence y = 92/19 Now substitute the value of y in Eqn 1 x + (7) x (92/19) = 33 x = 33 - 644/19 = -(17/19)
Eqn (A): => 2x + 5y = 16 Eqn (B): => 5x + 2y = -2 5*Eqn (A) - 2*Eqn (B): 21y = 84 => y = 4 Substituting for y in Eqn (a): x = -2
Eqn 1: x - 2y = 3 Eqn 2: 5x + 3y = 2 5 times Eqn 1: 5x - 10y = 15 Eqn 2 - 5*Eqn 1: 0x + 13y = -13 so that 13y = -13 or y = -1 Substitute in Eqn 1: x + 2*1 = 3 x + 2 = 3 x = 1 Solution: x = 1, y = -1
Photosynthesis equation: 6CO2 + 6H2O + light energy -> C6H12O6 + 6O2 Cellular respiration equation: C6H12O6 + 6O2 -> 6CO2 + 6H2O + ATP
First lets look at the equation for a circle... (x - h)2 + (y - k)2 = r2 What do I know? The center must be constant so H and K are two variables I need to solve for... and I know I need to solve for R as well! take your points (i'm making a simple circle with center (0,0) with radius of 1...) 1,0 0,1 -1,0 I have my 3 points.... I will make myself 3 equations: eqn #1a: (1 - h)2 +(0 - k)2 = r2 eqn #2a: (0 - h)2 +(1 - k)2 = r2 eqn #3a: (-1 - h)2 +(0 - k)2 = r2 Ok so now I have some equations to play with... eqn #1b: h2 - 2h +1 + k2 = r2 eqn #2b: h2 + k2-2k +1 = r2 eqn #3b: h2 + 2h +1 + k2= r2 The rest is subsitution... We want r's value so lets see. Lets take eqn #1b and #2b... h2 - 2h +1 + k2 = h2 + k2-2k +1 Balance it out cancle and such try to make it = 0 - 2h = -2k h = k so now I know I can replace h with k or k with h It doesn't matter which one I make to the other if we had 2h = k I would put 2h in its just simpler... keep in mind either way works. eqn #1c: k2 - 2k +1 + k2 = r2 eqn #2c: k2 + k2-2k +1 = r2 eqn #3c: k2 + 2k +1 + k2= r2 collect terms eqn #1c: 2k2 - 2k +1 = r2 eqn 1 is the same as eqn 2! eqn #2c: 2k2 - 2k +1 = r2 eqn #3c: 2k2 + 2k +1= r2 We really have two equations left now it doesn't always work out this way. Set eqn 2 and 3 equal to each other via the r 2k2 - 2k +1 = 2k2 + 2k +1 do some cancling -2k = 2k at this point I know k = 0 its the only viable answer.... take any one of our origonal equations.... h2 - 2h +1 + k2 = r2 all the parts with k or h = 0; remember h = k = 0 1 = r2 1 = r If I had time I would do a more complex equation for yea... but this kinda gets the point across...
6x - 5y = 11 --------- Eqn 1 -4x + 7y = 11 ------- Eqn 2 ___________ 24x - 20y = 44 ------ Eqn 1 multiplied by 4 on both sides, equation won't change -24x + 42y = 66 ---- Eqn 2 multiplied by 6 on both sides, equation won't change _____________ 0x + 22y = 110 ----- Adding the two eqns above gives you this result _____________ That is 22y = 110, or y = 5 (Answer) Now substitute the value of y in either eqn 1 or eqn 2 (we will use eqn 1) 6x - 5(5) = 11 6x = 11 + 25 = 36 x = 6 (Answer)
Vert is green.
Juan Vert's birth name is Juan Vert Carbonell.
Vert Ramp=a halfpipe with a vert on the top edge of both sides of it to enable the rider to catch air. (vert)icalQuarterpipe=1/2 of a vert ramp or halfpipe
The radius in the eqn. of capacitance is actual outside radius of the conductor whereas for inductance eqn. The radius is the self GMD of the conductor.
witch vert ramp?
Vert means turn