What we know so far:
Side 1 = 55m
Side 2 = 65m
Angle 1 = 40°
Angle 2 = 30°
What we are looking for:
Toby's Angle = ?
The distance x = ?
We need to look for Toby's angle so that we can solve for the distance x by assuming that the whole figure is a SAS (Side Angle Side) triangle.
Solving for Toby's Angle:
We know for a fact that the sum of all the angles of a triangle is 180°; therefore,
180° - (Side 1 + Side 2) = Toby's Angle
Toby's Angle = 180° - (40° + 30°)
Toby's Angle = 110°
Since we already have Toby's angle, we can now solve for the distance x by using the law of cosines r² = p²+ q²<span>− 2pq cos R where r is x, p is Side1, q is Side2, and R is Toby's Angle.
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x² = Side1² + Side2² - 2[(Side1)(Side2)] cos(Toby's Angle)
x² = 55² + 65² - 2[(55)(65)] cos(110°)
x² = 3025 + 4225 -7150[cos(110°)]
x² = 7250 - 2445.44
x = √4804.56
x = 69.31m
∴The distance, x, between two landmarks is 69.31m
Since were trying to solve for n, you have to isolate it so you would divide both sides by 10, and that will give you n = 4
Answer:
try and plug in number like 15 like and then find the answer that makes the 125
Step-by-step explanation:
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.
