Answer:
2
Step-by-step explanation:
Simplify the following:
(-(2 + 2/3))/(-(1 + 1/3))
(-(2 + 2/3))/(-(1 + 1/3)) = (-1)/(-1)×(2 + 2/3)/(1 + 1/3) = (2 + 2/3)/(1 + 1/3):
(2 + 2/3)/(1 + 1/3)
Put 1 + 1/3 over the common denominator 3. 1 + 1/3 = 3/3 + 1/3:
(2 + 2/3)/(3/3 + 1/3)
3/3 + 1/3 = (3 + 1)/3:
(2 + 2/3)/((3 + 1)/3)
3 + 1 = 4:
(2 + 2/3)/(4/3)
Put 2 + 2/3 over the common denominator 3. 2 + 2/3 = (3×2)/3 + 2/3:
((3×2)/3 + 2/3)/(4/3)
3×2 = 6:
(6/3 + 2/3)/(4/3)
6/3 + 2/3 = (6 + 2)/3:
((6 + 2)/3)/(4/3)
6 + 2 = 8:
(8/3)/(4/3)
Multiply the numerator by the reciprocal of the denominator, (8/3)/(4/3) = 8/3×3/4:
(8×3)/(3×4)
(8×3)/(3×4) = 3/3×8/4 = 8/4:
8/4
The gcd of 8 and 4 is 4, so 8/4 = (4×2)/(4×1) = 4/4×2 = 2:
Answer: 2
Answer:
x² -7x -18
Step-by-step explanation:
In algebra, a binomial consists solely of a sum or subtraction of two monomials.
To solve the multiplication of two binomials we have to multiply term by term and then simplify common terms.
(x + 2)(x - 9) = x(x) + x(-9) +2(x) + 2(-9) = x² -9x + 2x -18 = x² -7x - 18
Answer:
5 feet by 1 foot
took the unit test on edge
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>
