using the laws of indices, simplify the folowing. 1) 9^5/2 and 2)(s^3t^2)^4 also solve 6b+4/3b=41/5 solve for b.
2007-01-20 23:17:45 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1) 9^5/2 = sqrt 9 to the power of 5 = 9²\/9 = 81\/9
2)(s³t²)^4 = s¹²t^8
3)6b+4/3b=41/5
5(6b + 4) = 3b(41)
30b + 20 = 123b
30 b - 123b = -20
-93b = -20
b = 20/93
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2007-01-21 01:20:48 · answer #1 · answered by aeiou 7 · 5⤊ 0⤋
1.9^(5/2)=9^(2.5)=9^(2+.5)
={9^2}*{9^.5}=81*sq.root of 9
=81*(+ or - 3) i.e. -243 or +243.
2. s^3t^2=s^(3t*3t)=s^(9t^2).
Hence, the given expression
={s^(9t^2)}^4=s^{4*9t^2}=s^(36t+8).
3.Multiply both sides by b, to get
6b*b+(4/3)=41b/5.
Multiply both sides by 15 to get
90b^2+20=123b or
90b^2-123b+20=0. Solving,
b=123(plus or minus 89.04)*(1/180)
or b=(123+89.04)/180 or (123-89.04)/180
(correct to two decimals)
i.e. b=1.18 or 0.19
2007-01-21 00:47:34 · answer #2 · answered by greenhorn 7 · 0⤊ 1⤋
1)=3^5=243
2)s^12t^8
3) 18b^2+4=123b/5
90b^2-123b+20=0
b=0.19or =1.18
2007-01-20 23:49:10 · answer #3 · answered by Maths Rocks 4 · 0⤊ 1⤋
gosh that's a tough one, I'd reckon it to be x or x.5!!!
sorry just being silly
2007-01-21 01:20:13 · answer #4 · answered by Anonymous · 1⤊ 1⤋
1) 4.5^3 (I think)
2007-01-20 23:24:22 · answer #5 · answered by don't stop the music ♪ 6 · 0⤊ 1⤋