Last Name 1
Student’s Name
Professor’s Name
Subject
Date
Task 8
1
Using Heron’s formulae
Area = √(s(s-a) x (s-b) x (s-c))
S = (a+b+c)/2 = (6+9+c)/2 = (15+c)/2=7.5+0.5c
Area = 18cm2
18 = √((7.5+0.5c) x (7.5+0.5c-6) x (7.5+0.5c-9) x (7.5+0.5c-c))
182 = (√((7.5+0.5c) x (1.5+0.5c) x (0.5c-1.5) x (7.5-0.5c)))2
324 = (7.5+0.5c) x (7.5-0.5c) x (1.5+0.5c) x (0.5c-1.5)
324 = (56.25-0.2c2) x (0.25c-2.25)
Using μ substitution then quadratic formula;
c = ±√197.49844718 = 14.05cm
or ± √36.50155281 = 6.04cm
Using cosine rule to find the value of ∠B;
a2= b2+ c2-2bc cosA
When c = 14.05 cm
14.052 = 62 + 92 - (2x6x9)cosB
80.4025= -108 cosB
cosB = 80.4025/-108 = -0.7445
cos-1 -0.7445 = 138.11°
Last Name 2
when c = 6.04 cm
6.042 = 62 + 92 - (2x6x9) cosB
-80.5184 = -108 cosB
cosB = -80.5184/-108 = 0.7455
cos-1 0.7455 = 41.79°
∠B = 41.79° and 138.11°
2
tan30°(6sin135°+4cos210°)
sin135° = sin (90°+45°)
cos210°=cos (180°+30°)
tan30°=1/√3 = √3/3
sin90° = 1;
sin45° = cos45° = 1/(√2)
cos180°= -1
cos30°=√3/2
sin(A+B) = sinAcosB + sinBcosA
sin (90°+40°) = sin90°.cos45° + sin45°.cos90° = 1.cos45°+sin45°.0 = cos45° = 1/√2 = √2/2
cos (A+B) = cosAcosB - sinAsinB
cos (180°+30°) = cos180°.cos30° - sin180°.sin30° = -1.cos30°- sin30°.0 = -cos30° = -√3/2
tan30°(6sin135°+4cos210°) = √3/3 x ((6 x √2/2) + (4 x -√3/2)) = √3/3 x (3√2) + √3/3 x (2√3)
= -2 + √6
3
Using LHS to prove RHS
Last Name 3
sin3θ (1 - cos2θ) ≈ 6θ3
For small θ;
Sin θ ≈ θ
cosθ = 1 - 2 sin2(θ/2)
cosθ ≈ 1 - 2(θ/2)2 = 1 - θ2/2
Cos 2θ = 1 - 2θ2/2 = 1 - 4θ2/2 = 1-2θ2
As 3θ° tends to 0°; sin3θ becomes 3θ
3θ(1-(1-2θ2)) = 3θ(2θ2) = 6θ3
4
cosθ (3sin2θ + 4 cos2θ ) ≡ 3cosθ+cos3θ
Using the LHS to prove RHS;
Replacing cosθ = sinθ/tanθ
cosθ (3sin2θ + 4 cos2θ)
Using commutative property of multiplication and adding the exponentials,
3cosθ.sin2θ + 4 cos3θ
sin2θ + cos2θ = 1
sin2θ = 1 - cos2θ
3cosθ.sin2θ + 4 cos3θ
3cosθ (1 - cos2θ) + 4 cos3θ
3cosθ - 3cos3θ + 4 cos3θ = 3 cosθ + cos3θ
5
(36+4.27÷4-18) ÷5
Solving the terms in the brackets then outside the brackets;
Last Name 4
Solve the terms with Division, multiplication, addition, and then subtraction.
(36 + 4x27/4-18) ÷ 5
(36 + 4 x 6.75 - 18) ÷ 5
(36 + 27 - 18) ÷ 5
(63 - 18) ÷ 5
45 ÷ 5 = 9
6
Training Heart Rate = ((maximum heart rate – resting heart rate) × 0.65) + resting heart rate
maximum heart rate = 220 – person’s age
Maximum Heart Rate = 220 – 50 = 170
Training Heart Rate = ((170 - 65) x 0.65) + 65 = 105 x 0.65 + 65 = 68.25 + 65 = 133.25
For complete beats per minute(bpm);
Training heart Rate = 133bpm
7
Let the “a number” be N
“Five times a Number, N” will be 5N
The sum of “five times a Number” plus three will be 5N+3
= 5N+3
8
4(x + 1) = -15
4x + 4 = -15
4x = -15 - 4
4x = -19
Last Name 5
x = -19/4
x = -4.75
9
The magnitude of y+7 will be the absolute value
The absolute value is the distance between a number and Zero, 0.
If y = -2;
Replacing y with-2 we will have;
|-2 + 7|
We will move to 5
|-2+7| = 5