2 more questions thanks

Answers

Answer 1

Answer: 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.

Answer 2

Answer: Hello,
Please, see the detailed solution in the attached files.
Thanks.

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Answers

Answer:

(4, 5)

Step-by-step explanation:

Dilation by factor d about point (a, b) effects the transformation ...

(x, y) ⇒ d((x, y) -(a, b)) +(a, b)

⇒ (d(x-a) +a, d(y-b) +b)

(x, y) ⇒ (dx +a(1-d), dy +b(1-d))

We have (x, y) = (3, 4) and (a, b) = (1, 2) and d=3/2. Filling in these numbers, we get ...

(3, 4) ⇒ ((3/2)(3)+1(1-3/2), (3/2)(4)+2(1-3/2)) = (9/2 -1/2, 6 -1) = (4, 5)

Point E' is (4, 5).

_____

If you look at the graph, you realize that E is (2, 2) from (1, 2). Dilation by a factor of 3/2 moves it to (3/2)(2, 2) = (3, 3) from (1, 2). So ...

E' = (3+1, 3+2)

E' = (4, 5)

Describe in words the region of R^3 represented by inequality 1<= x^2 + y^2 + z^2 <= 5

Answers

The region is bounded by 2 spheres, both centered at the origin and having radii 1 and 5, respectively.

Which lists all the integer solutions of the inequality |x| < 3?A. 0, 1, and 2
B. 0, 1, 2, and 3
C. –2, –1, 0, 1, and 2
D. –3, –2, –1, 0, 1, 2, and 3

Answers

If you would like to know which lists all the integer solutions of the inequality |x| < 3, you can calculate this using the following steps:

|x| < 3
- 3 < x < 3
x = {-2, -1, 0, 1, 2}

The correct result would be C. -2, -1, 0, 1, and 2.

simplify complex fraction

3/4 + 1/3/1/6 + 3/4

Answers

= (3/4 + 1/3)/(1/6 + 3/4)
= (3/4 + 1/3)/(1/6 + 3/4)
Multiply numerator and denoiminator by 12
=12(3/4+1/3) / 12(1/6 + 3/4)
= (3*3 + 1*4) / (2 + 3*3)
= 13/11

In △PQR, if the measures of ∠Q and ∠R are 100° and 45°respectively, find the measure of ∠P