The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has length 2. What is the area of the shaded sector formed by obtuse angle WXY?
RADIUS = 2
CHORD = 2
RADIUS --> XY , XZ , WX
( BEZ THEY TOUCH CIRCUMFERENCE OF THE CIRCLES AFTER STARTING FROM CENTRE OF THE CIRCLE)
THE AREA OF THE SHADED SECTOR FORMED BY OBTUSE ANGLE WXY.
AREA COVERED BY THE ANGLE IN A SEMI SPHERE
Total Area Of The Semi Sphere:-
Area Under Unshaded Part .
Given a triangle with each side 2 units.
This proves that it's is a equilateral triangle which means it's all angles r of 60° or π/3 Radian
So AREA :-
Total Area - Area Under Unshaded Part
Do you mean a_(n+1), worded a sub (n+1)?
If so yes. If the function of the sequence is getting smaller or more negative with each term.
Answer:
the two angles are 55° each
Step-by-step explanation:
Let the two equal angles be y each
Therefore,
70° + y + y = 180°
Collect like terms
2y = 180 — 70
2y = 110
Divide both side by the coefficient of y i.e 2
y = 110/2
y = 55°
Therefore, the two angles are 55° each
You should have an angle of elevation of 30 degrees.
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
