Is it 1/2×b×h
to find the area
or
base×height
Answer:
Im not fully sure but I think it is negative 2 and negative 3
He will have to take out 20 hectograms in order to meet the weight limit.
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>
Answer/Step-by-step explanation:
Equation to represent the daily rental cost for each type of truck can be written as follows:
Daily rental cost for Trucks-A-Lot = 42 + 0.72m
Daily rental cost for Move-in-Truckers = 70 + 0.12m
Where, m = Emily's mileage
To determine the number of miles for which the truck cost the same amount, set both equations equal to each other and solve for m.

Collect like terms


Divide both sides by 0.6


At approximately 47 miles, both trucks would cost the same amount.
Check:
Daily rental cost for Trucks-A-Lot = 42 + 0.72m
Plug in the value of x = 47
= 42 + 0.72(47) = $75.84 ≈ $76
Daily rental cost for Move-in-Truckers = 70 + 0.12m
Plug in the value of x = 47
= 70 + 0.12(47) = $75.64 ≈ $76