Answer:
a) $1056.33 b) 23 years
Step-by-step explanation:
a) 10000(1+1.85/100)^3=10565.33 (2d.p.)
b) let x be the no. of years
15000 = 10000(1+1.85/100)^x
1.5 = 1.0185^x
ln both sides
ln 1.5 = x ln 1.0185
x = ln 1.5/ln 1.0185
=22.11
=23 years (rounded up)
You would have to add a positive 6 to the negative 6 to get zero. Lets say you have -2 in order to get it to zero or just any positive number, you have to add a positive of the same value or higher to be able to get it there. I hope you understand that.
Answer:
- (6-u)/(2+u)
- 8/(u+2) -1
- -u/(u+2) +6/(u+2)
Step-by-step explanation:
There are a few ways you can write the equivalent of this.
1) Distribute the minus sign. The starting numerator is -(u-6). After you distribute the minus sign, you get -u+6. You can leave it like that, so that your equivalent form is ...
(-u+6)/(u+2)
Or, you can rearrange the terms so the leading coefficient is positive:
(6 -u)/(u +2)
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2) You can perform the division and express the result as a quotient and a remainder. Once again, you can choose to make the leading coefficient positive or not.
-(u -6)/(u +2) = (-(u +2)-8)/(u +2) = -(u+2)/(u+2) +8/(u+2) = -1 + 8/(u+2)
or
8/(u+2) -1
Of course, anywhere along the chain of equal signs the expressions are equivalent.
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3) You can separate the numerator terms, expressing each over the denominator:
(-u +6)/(u+2) = -u/(u+2) +6/(u+2)
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4) You can also multiply numerator and denominator by some constant, say 3:
-(3u -18)/(3u +6)
You could do the same thing with a variable, as long as you restrict the variable to be non-zero. Or, you could use a non-zero expression, such as 1+x^2:
(1+x^2)(6 -u)/((1+x^2)(u+2))
