Divide the figure into two parts, a triangle and a square
TRIANGLE
the base (b) = 5 cm
the height (h) = 4 cm
Calculate the area of the triangle
a = 1/2 × b × h
a = 1/2 × 5 × 4
a = 20/2
a = 10
The area of the triangle is 10 cm²
SQUARE
the side (s) = 5 cm
Calculate the area of the square
a = s²
a = 5²
a = 25
The area of the square is 25 cm²
THE AREA OF THE FIGURE
a figure = a of triangle + a of square
a figure = 10 + 25
a figure = 35
The area of the figure is 35 cm²
Let ∆ABC is a triangle such that side length of AB is c , BC is a and CA is b .
Given, AB = c = 8
m∠A=60°
m∠C=45°
m∠B = (180 - 45 - 60)° = 75°
use sine rule to get b and c
\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}asinA=bsinB=csinC
so, \frac{sin60^{\circ}}{8}=\frac{sin45^{\circ}}{c}=\frac{sin75^{\circ}}{b}8sin60∘=csin45∘=bsin75∘
sin60°/8 = sin45°/c
(√3/2)/8 = (1/√2)/c
√3/16 = 1/√2c => c = 16/√6
also, sin60°/8 = sin75°/b
b = 8sin75°/sin60°
= {8 × (√3 + 1)/2√2}/{√3/2}
= 4√2(√3 + 1)/√3
hence, perimeter of ∆ABC = a + b + c
= 8 + 16/√6 + 4√2(√3 + 1)/√3
= 8 + (16 + 8√3 + 8)/√6
= 8 + (24 + 8√3)/√6
= 8 + 4√6 + 4√2
area of ∆ABC = 1/2 absinC
= 1/2 × 8 × 4√2(√3 + 1)/√3 × sin45°
= 4 × 4√2(√3 + 1)/√3 × 1/√2
= 16(√3 + 1)/√3
= (48 + 16√3)/3
= 16 + 16/√3
Remark
The closest answer is D, but the answer has to be rewritten: f(x) = - 2/3 (6)^(x - 1) where x is 1,2,3,4,5,6 .... to plus infinity. If that is not what you are given, there is no answer.
Discussion
The minus sign is misplaced. There are 2 reasons.
1. When x is odd, then the minus sign is not there and every other term will be plus.
2. For a related reason, you cannot get a minus 2/3 if the sign is affected by the power. The first term must be -2/3. (-6)^0 power is 1 not -1
Answer: There is no answer, but you could try C if D does not work.
25a + 5b - 13
5(5a + b) - 13 is your answer
using the distributive property, we can distribute 5 to all terms within the parenthesis:
5(5a + b) - 13 = 25a + 5b - 13
5(5a) = 25a
5(b) = 5b
25a + 5b - 13 = 25a + 5b - 13 (True)
hope this helps