Here’s a pic of my work, it shows that the numbers 2,7,0 repeat.
Hope this helps!! :)
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ
Use the Sum/Difference Identities:
sin(α + β) = sinα · cosβ + cosα · sinβ
cos(α - β) = cosα · cosβ + sinα · sinβ
Use the Unit circle to evaluate: sin45 = cos45 = √2/2
Use the Double Angle Identities: sin2Ф = 2sinФ · cosФ
Use the Pythagorean Identity: cos²Ф + sin²Ф = 1
<u />
<u>Proof LHS → RHS</u>
LHS: 2sin(45 + 2A) · cos(45 - 2A)
Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)
Unit Circle: 2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]
Expand: 2[(1/2)cos²2A + cos2A·sin2A + (1/2)sin²2A]
Distribute: cos²2A + 2cos2A·sin2A + sin²2A
Pythagorean Identity: 1 + 2cos2A·sin2A
Double Angle: 1 + sin4A
LHS = RHS: 1 + sin4A = 1 + sin4A 
Answer:
(
−
6
,
−
30
)
Step-by-step explanation:
Rewrite in vertex form and use this form to find the vertex (
h
,
k
)
.
Answer:
% Po lost = 100[1 - e^(-0.005t)] %; 73.0 g
Step-by-step explanation:
p(t) = 100e^(-0.005t)
Initial amount: p(0) = 100
Amount remaining: p(t) = 100e^(-0.005t)
Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]
% of Po lost = amount lost/initial amount × 100 %
= [1 - e^(-0.005t)] × 100 % = 100[1 - e^(-0.005t)] %
p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g
The mass of polonium remaining after 63 days is 73 g.