PLEASE HELP ME WITH THIS!!!!
1 answer:
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Since the big triangle is an isosceles one, we conclude that the angles of the left triangle are: 32, 90 and 3x-10. Sinche the sum of this angles have to be 180° we have:
Solving for x, we have:
Therefore, x=22.67 and the answer is A.
Answer: The correct option is A, itis the product of the initial population and the growth factor after h hours.
Explanation:
From the given information,
Initial population = 1000
Increasing rate or growth rate = 30% every hour.
No of population increase in every hour is,
Total population after h hours is,
It is in the form of,
Where is the initial population, r is increasing rate, t is time and [tex(1+r)^t[/tex] is the growth factor after time t.
In the above equation 1000 is the initial population and is the growth factor after h hours. So the equation is product of of the initial population and the growth factor after h hours.
Therefore, the correct option is A, itis the product of the initial population and the growth factor after h hours.
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