<span>Given a circle with radius r = 8 and a sector with subtended angle measuring 45°, find the area of the sector and the arc length.
They've given me the radius and the central angle, so I can just plug into the formulas. For convenience, I'll first convert "45°" to the corresponding radian value of π/4:
A = ((pi/4)/2)(8^2) = (pi/8)(8^2) = 8pi, s = (pi/4)(8) = 2pi
area A = 8π, arc-length s = 2π
Given a sector with radius r = 3 and a corresponding arc length of 5π, find the area of the sector.
For this exercise, they've given me the radius and arc length. From this, I can work backwards to find the subtended angle.
Then I can plug-n-chug to find the sector area.
5pi = (theta)(3), (theta) = (5/3)pi
So the central angle is (5/3)π.
Then the area of the sector is:
A = ((5/3)pi / 2)*(3^2) = ((5/6)pi)*(9) = (15/2)pi
A = (15 pi) / 2
</span><span>90/360 = 0.25 pi sq ft</span>
|x| this means absolute value making what ever number in the inside positive UNLESS you have a negative out front. Ex |8|=8. -|8|=-8. |-8|=8
Answer:
1022.50
Step-by-step explanation:
1000*.03= 30. (12mos interest)
30/4=7.50 quarterly interest
7.50x3 (9mos)= 22.50 interest
The descriminant is b^2-4ac so for 26 it would be -5^2 - 4 (1) (7). The quadratic formula is -B plus or minus the square root of b squared minus 4ac over 2a
Answer:
8978 grams
Step-by-step explanation:
The equation to find the half-life is:
N(t) = amount after the time <em>t</em>
= initial amount of substance
t = time
It is known that after a half-life there will be twice less of a substance than what it intially was. So, we can get a simplified equation that looks like this, in terms of half-lives.
or more simply 
= time of the half-life
We know that = 35,912, t = 14,680, and =7,340
Plug these into the equation:
Using a calculator we get:
N(t) = 8978
Therefore, after 14,680 years 8,978 grams of thorium will be left.
Hope this helps!! Ask questions if you need!!
