When they say to solve for a variable, they want you to isolate/get the variable by itself.
5.) Solve for b (get b by itself)
Multiply 2 on both sides to get rid of the 1/2
2A = bh Divide h on both sides
6.) Solve for r
Divide 2 and 
on both sides
Answer:
Step-by-step explanation:
The area of a pie-slice shape can be found with the equation 
where n is the angle of the shape and r is the radius of the circle. In this case, n=60 and r=6. Therefore, the total area of the pie-slice is 
Next, the triangle that makes up the unshaded part of the segment is an equilateral triangle because the angle is 60. The area of an equilateral triangle is 
where r is the length of one of its side (in this case, r=6). Plugging that in, we get an area of 
Finally, the area of the shaded is the area of the pie-slice minus the area of the triangle:
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.
