Your pay is described by a piecewise function p(t) where t is hours worked.

The domain of this function is the number of hours you can work in a week—zero to 168. The range is the set of values p(t) can take on, which we have said is p(0) to p(168). This problem, however, is asking for the range for a typical 40-hour week. That will be p(0) to p(40):
the interval [0, 280].
If <em>x</em> + 1 is a factor of <em>p(x)</em> = <em>x</em>³ + <em>k</em> <em>x</em>² + <em>x</em> + 6, then by the remainder theorem, we have
<em>p</em> (-1) = (-1)³ + <em>k</em> (-1)² + (-1) + 6 = 0 → <em>k</em> = -4
So we have
<em>p(x)</em> = <em>x</em>³ - 4<em>x</em>² + <em>x</em> + 6
Dividing <em>p(x)</em> by <em>x</em> + 1 (using whatever method you prefer) gives
<em>p(x)</em> / (<em>x</em> + 1) = <em>x</em>² - 5<em>x</em> + 6
Synthetic division, for instance, might go like this:
-1 | 1 -4 1 6
... | -1 5 -6
----------------------------
... | 1 -5 6 0
Next, we have
<em>x</em>² - 5<em>x</em> + 6 = (<em>x</em> - 3) (<em>x</em> - 2)
so that, in addition to <em>x</em> = -1, the other two zeros of <em>p(x)</em> are <em>x</em> = 3 and <em>x</em> = 2
Answer:
The fish is closer to the surface of the water because |+4| = 4 and |−3| = 3, and 3 < 4.
Step-by-step explanation:
Answer:
irrational
Step-by-step explanation:
a rational number is always a whole number
Lets solve all of these:-
#1
√361 = 361 · 2
?
√361 = 361 · 2
√361 = 19
361 · 2 = 722
19 ≠ 722
So this equation is not true
#2:-
√361 = 19²
?
√361 = 19²
√361= 19
19² = 19 · 19 = 361
19 ≠ 361
So this equation is not true
√361 = 361 ÷ 2
?
√361 = 361 ÷ 2
√361 = 19
361 ÷ 2 = 180.5
√361 ≠ 361 ÷ 2
So this equation is not true
√361 = √19²
√361 = 19
√19² = 19
19 = 19
SO the last one is right. Hope I helped ya!! xD