ANSWER:
[a] Chord –» AE.
[b] Minor arc –» arcAB.
[c] Radius –» BC.
[d] Major arc –» arcBAE.
[e] Tangent line –» DE.
[f] Inscribed angle –» ∠BAE.
[g] Central angle –» ∠BCE.
[h] Secant line –» AB.
This is a combination of the formula of several figures.
The first one is the bottom of the cylinder AKA a circle.
The second one is the cylinder - the bottom and the top.
The third one is a cone without a bottom.
The first one can be written like this: r^2*3.14 where r is the radius of the circle
The second one is written like this: 2*3.14*r*h where r is the radius, and h the height
The third one is written like this: l*r*3.14
After this you just have to input the values.
bottom cylinder Cone (Use calculator if possible)
(3^2*3.14)+(2*3.14*3*5)+(6*3*3.14)=
28.26 + 94.2 + 56.52 = 178.98 in^2
The correct answer is 178.98 in^2.
Youre welcome.
EXTRA:
If youre wondering how to find the different formulas, you can search for the full formula, then remove the parts you do not need, by matching them with the formula of the part you want to remove.
Example:
Cylinder:
SA = 2(pi<span> r</span><span> 2</span><span>) + (2 </span><span>pi </span><span>r)* h
</span>In this case, i didnt need the top and bottom, then i look for the formula of a circle.
Circle SA = pi r^2
Is this anywhere in the cylinder? Yes! The first part.
Then we are left with:
2 + (2 pi r)*h
But that 2 looks strangely placed, and with some reason, one quickly understands that it means "top AND bottom", but since we removed the circle surface, we have to remove that part too.
So the final result of the formula is:
2 pi r h
Answer:
60
Step-by-step explanation:
1/3 of all attempts expected to be foul shots
1/3*90= 30 fouls
90-30= 60 success
200/5 = 40 mph 200/4 = 50 mph Let x be the speed in still waters and y the speed of current. x + y = 50 x - y = 40 ADD 2x = 90 x = 45 mph ANSWER is C) 2) C
Answer:
Step-by-step explanation:
Volume of water in the tank = 1000 L
Let y(t) denote the amount of salt in the tank at any time t.
Initially, the tank contains 60 kg of salt, therefore:
y(0)=60 kg
<u />
<u>Rate In</u>
A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.
=(concentration of salt in inflow)(input rate of solution)
<u>Rate Out</u>
The solution is mixed and drains from the tank at the same rate.
Concentration, 
=(concentration of salt in outflow)(output rate of solution)
Therefore, the differential equation for the amount of Salt in the Tank at any time t:
