factor out the common monomial, 5.
5(r^3-1)
Factor as a difference of cubes.
5(r-1)(r^2+r+1)
(r + 1)(3r - 5)
1,5
1 and 5
P=I^2R. 80^2=6400*2=12,800 Watts. That is for D.C. For A.C. you would use the R.M.S. or average wattage.
The factors of 3 are 1 and 3 The factors of 12 are 1, 2, 3, 4, 6, and 12 The factors of 18 are 1, 2, 3, 6, 9, and 18
The GCF is 1.
2^4 x 7
to put out the power fector you have to divided apparent power with true power.AnswerYou can determine the true power of any load using a wattmeter. To find the apparent power, you use a voltmeter to measure the supply voltage and an ammeter to measure the load current, and multiply the two readings together.If you then want to go on to find the power factor, then you divide the true power by the apparent power. If you want to find the reactive power you use the following equation:(reactive power)2 = (true power)2 x (apparent power)2
If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2
If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2
3r+10=5r-46 --> add 46 to both sides 3r+10+46=5r-46+46 3r+56=5r --> subtract 3r from both sides 3r+56-3r=5r-3r 56=2r --> divide both sides by 2 56/2=2r/2 28=r
It is not correct to say that 8 plus 2 8r-2r 4 3r plus 2 r = 0
(3r - 2)(3r - 2) or (3r - 2) squared.
r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)
(3r + 2)(r - 5)
The radius is tripled, too. Suppose: (1) C = 2πr (2) C' = 2πr' (3) C' = 3C Then C' = 3(2πr) = 2π(3r) But C' = 2πr', so 2πr' = 2π(3r) and finally r' = 3r.
(2S,3S)-2,3-Pentanediol and (2R,3R)-2,3-Pentanediol
1/2r
1)5r+7=13+2r+3r 2)5r+7=13+5r 3)5r's cancel 7=13 4)Subtract thirteen on both sides 7 =13 -13 -13 5)Answer is no solution because -6 doesn't equal to 0. -6=0 NO SOLUTION!!! Have fun with math!
Sodium chloride is NaCl.Sodium hyaluronate has a very long and complicate name:Sodium (2S,3S,4R,5R,6R)-3- [(2S,3R,4R,5S,6R)-3-acetamido-4-[(2R,3R,4S,5S,6S)-6-carboxy-3,4,5-trihydroxyoxan-2-yl]oxy-5-hydroxy-6-(hydroxymethyl)oxan-2-yl]oxy-6-[(2R,3R,4R,5S,6R)-3-acetamido-2,5-dihydroxy-6-(hydroxymethyl)oxan-4-yl]oxy-4,5-dihydroxyoxane-2-carboxylic acid.