Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS =
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
Let
x-------> the number of dinner
y-------> the number of lunch
we know that
-------> equation A
------> equation B
Substitute equation B in equation A
![8[y]+5y \leq 42](https://tex.z-dn.net/?f=8%5By%5D%2B5y%20%5Cleq%2042)



so
the greatest number of lunch is 

Hence
the greatest number of dinner is 
therefore
the greatest number of meals is

<u>the answer is</u>

Answer:
821.94
Step-by-step explanation:
Answer:
£1999.85
Step-by-step explanation:
A = P(1 + rt)
Where:
A = Amount after t years
P = Initial amount invested = £1850
r = Interest rate = 2.7%
t = Time in years = 3 years
Calculation:
First, converting R percent to r a decimal
r = R/100 = 2.7%/100 = 0.027 per year.
Solving our equation:
A = £1850(1 + (0.027 × 3))
A = £1999.85
Therefore, Brian will have £1999.85 after 3 years.