Answer:
B. 0.73
Step-by-step explanation:
Add all numers so the first is 2/10 = 0.2
then 35/100 = 0.35
and then 16/100 = 0.16
finally 2/100 = 0.02
Now add all decimals which is equal to 73/100 = 0.73 and that's your answer.
^Virtuoso^
$350= $130+$20x, x= amount of weeks
$350-$130=$220 $20*11=220
Answer: 11 weeks
Answer:
B. 3/4
Step-by-step explanation:
5/8 Get a common denominator of 24
5/8 *3/3 = 15/24
A. 12/24 this is less than
B. 3/4 *6/6 = 18/24 greater than
C. 15/24 same value so equal to
D.4/12 * 2/ = 8/24 less than
Answer:
Irrational numbers cannot be written as a fraction, or the ratio of two whole numbers.
You would have to find a GCF (greatest common factor) and multiply it out of the radical.
ex.
= 2 x 2 x 2
when you have to multiplier numbers then you can combine two of them to take outside the radical, meaning 2x2x2
2
so would be closest to the positive 2 on the number line and 2 notches.
is an irrational number, so it might fall into 3.3 if turned into a decimal.
Step-by-step explanation:
There are two of them.
I don't know a mechanical way to 'solve' for them.
One can be found by trial and error:
x=0 . . . . . 2^0 = 1 . . . . . 4(0) = 0 . . . . . no, that doesn't work
x=1 . . . . . 2^1 = 2 . . . . . 4(1) = 4 . . . . . no, that doesn't work
x=2 . . . . . 2^2 = 4 . . . . . 4(2) = 8 . . . . . no, that doesn't work
x=3 . . . . . 2^3 = 8 . . . . . 4(3) = 12 . . . . no, that doesn't work
<em>x=4</em> . . . . . 2^4 = <em><u>16</u></em> . . . . 4(4) = <em><u>16</u></em> . . . . Yes ! That works ! yay !
For the other one, I constructed tables of values for 2^x and (4x)
in a spread sheet, then graphed them, and looked for the point
where the graphs of the two expressions cross.
The point is near, but not exactly, <em>x = 0.30990693...
</em>If there's a way to find an analytical expression for the value, it must involve
some esoteric kind of math operations that I didn't learn in high school or
engineering school, and which has thus far eluded me during my lengthy
residency in the college of hard knocks.<em> </em> If anybody out there has it, I'm
waiting with all ears.<em>
</em>
