9514 1404 393
Answer:
no feasible solutions
Step-by-step explanation:
There are so many inequalities that determining the area covered by all 5 of them is problematic. Hence, we have reversed all of them in the attached graph, so any feasible solution region would remain white. Alas, there is no such region.
This system of inequalities has <em>no feasible solution</em>.
Answer:
either of {-9.5, -3.5, 2.5} or {7, 13, 19}
Step-by-step explanation:
Let x represent the first number of the arithmetic sequence. Then the second number is x+6, and the third number is x+12. The geometric sequence is then ...
(x -4), (x +5), 3(x+9)
The common ratio is then ...
(x +5)/(x -4) = 3(x +9)/(x +5)
Cross-multiplying gives ...
(x +5)² = 3(x -4)(x +9)
x² +10x +25 = 3x² +15x -108 . . . . eliminate parentheses
0 = 2x² +5x -133 . . . . . . . . . . . . . . subtract the left side
0 = (2x +19)(x -7) . . . . . . . . . . . . . . factor
x = -9.5 . . or . . 7
The possible original three numbers are ...
{-9.5, -3.5, 2.5} . . or . . {7, 13, 19}
IL is a diameter, which is a straight line, which is equal to 180 degrees.
The arc of IL would be 180 degrees
You are told the arc of KL is 30 degrees.
The major arc of ILK would be 180 + 30 = 210 degrees.
These are two questions and two answers.
1) Problem 17.
(i) Determine whether T is continuous at 6061.
For that you have to compute the value of T at 6061 and the lateral limits of T when x approaches 6061.
a) T(x) = 0.10x if 0 < x ≤ 6061
T (6061) = 0.10(6061) = 606.1
b) limit of Tx when x → 6061.
By the left the limit is the same value of T(x) calculated above.
By the right the limit is calculated using the definition of the function for the next stage: T(x) = 606.10 + 0.18 (x - 6061)
⇒ Limit of T(x) when x → 6061 from the right = 606.10 + 0.18 (6061 - 6061) = 606.10
Since both limits and the value of the function are the same, T is continuous at 6061.
(ii) Determine whether T is continuous at 32,473.
Same procedure.
a) Value at 32,473
T(32,473) = 606.10 + 0.18 (32,473 - 6061) = 5,360.26
b) Limit of T(x) when x → 32,473 from the right
Limit = 5360.26 + 0.26(x - 32,473) = 5360.26
Again, since the two limits and the value of the function have the same value the function is continuos at the x = 32,473.
(iii) If T had discontinuities, a tax payer that earns an amount very close to the discontinuity can easily approach its incomes to take andvantage of the part that results in lower tax.
2) Problem 18.
a) Statement Sk
You just need to replace n for k:
Sk = 1 + 4 + 7 + ... (3k - 2) = k(3k - 1) / 2
b) Statement S (k+1)
Replace
S(k+1) = 1 + 4 + 7 + ... (3k - 2) + [ 3 (k + 1) - 2 ] = (k+1) [ 3(k+1) - 1] / 2
Simplification:
1 + 4 + 7 + ... + 3k - 2+ 3k + 3 - 2] = (k + 1) (3k + 3 - 1)/2
k(3k - 1)/ 2 + (3k + 1) = (k + 1)(3k+2) / 2
Do the operations on the left side and you will find it can be simplified to k ( 3k +1) (3 k + 2) / 2.
With that you find that the left side equals the right side which is a proof of the validity of the statement by induction.
