Answer:
A = 20sinθ(6 + 5 cosθ) cm²
Step-by-step explanation:
Drop perpendiculars DE and CF to AB.
Then, we have congruent triangles ADE and BCF, plus the rectangle CDEF.
The formula for the area of the trapezium is
A = ½(a + b)h
DE = 10sinθ
AE = 10cosθ
BF = 10cosθ
EF = CD = 12 cm
AB = AE + EF + BF = 10cosθ + 12 + 10 cosθ = 12 + 20cosθ
A = ½(a + b)h
= ½(12 +12 + 20 cosθ) × 10 sinθ
=(24 + 20 cosθ) × 5 sinθ
= 4(6 + 5cosθ) × 5sinθ
= 20sinθ(6 + 5 cosθ) cm²
Answer:
x = 6
Step-by-step explanation:
Given that F is between E and G , then
EF + FG = EG , substitute values
2x - 2 + 6x = 6x + 10 , that is
8x - 2 = 6x + 10 ( subtract 6x from both sides )
2x - 2 = 10 ( add 2 to both sides )
2x = 12 ( divide both sides by 2 )
x = 6
Recall that for all t,
cos²(t) + sin²(t) = 1
Now,
x = 5 cos(t) - 7 ⇒ (x + 7)/5 = cos(t)
y = 5 sin(t) + 9 ⇒ (y - 9)/5 = sin(t)
so that substituting into the identity above, we get
((x + 7)/5)² + ((y - 9)/5)² = 1
which we can rewrite as
(x + 7)²/25 + (y - 9)²/25 = 1
(x + 7)² + (y - 9)² = 25
and this is the equation of a circle centered at (-7, 9) with radius 5.
Answer:
tbh i be wasting m time and ur tiem no kap
Step-by-step explanation:
Answer:
There can be at most 
attendees in a corporate team-building event.
Step-by-step explanation:
Let 
denotes number of attendees in a corporate team-building event.
Fixed cost = $19
Cost charged per attendee = $1
Budget for the corporate team-building event = $31
Therefore,
So, there can be at most attendees in a corporate team-building event.
