The answer in itself is 1/128 and here is the procedure to prove it:
cos(A)*cos(60+A)*cos(60-A) = cos(A)*(cos²60 - sin²A)
<span>= cos(A)*{(1/4) - 1 + cos²A} = cos(A)*(cos²A - 3/4) </span>
<span>= (1/4){4cos^3(A) - 3cos(A)} = (1/4)*cos(3A) </span>
Now we group applying what we see above
<span>cos(12)*cos(48)*cos(72) = </span>
<span>=cos(12)*cos(60-12)*cos(60+12) = (1/4)cos(36) </span>
<span>Similarly, cos(24)*cos(36)*cos(84) = (1/4)cos(72) </span>
<span>Now the given expression is: </span>
<span>= (1/4)cos(36)*(1/4)*cos(72)*cos(60) = </span>
<span>= (1/16)*(1/2)*{(√5 + 1)/4}*{(√5 - 1)/4} [cos(60) = 1/2; </span>
<span>cos(36) = (√5 + 1)/4 and cos(72) = cos(90-18) = </span>
<span>= sin(18) = (√5 - 1)/4] </span>
<span>And we seimplify it and it goes: (1/512)*(5-1) = 1/128</span>
Answer:
5
Step-by-step explanation:
49/9 = 5
Answer:
the area is 30 correct me if Im wrong
Step-by-step explanation:
I honestly don’t know the equation but I just took the starting price, $230,000 and multipled it by .03 which was 6,900 and subtracted that from 230,000: 223,190. I repeated that 5 times. The answer I got is 197,508.83
Answer: The total interest paid on the mortgage is $179550
Step-by-step explanation:
The initial cost of the property is $300000. If he deposits $30000, the remaining amount would be
300000 - 30000 = $270000
Since the remaining amount was compounded, we would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = 270000
r = 2% = 2/100 = 0.02
n = 12 because it was compounded 12 times in a year.
t = 25 years
Therefore,
A = 270000(1+0.02/12)^12 × 25
A = 270000(1+0.0017)^300
A = 270000(1.0017)^300
A = $449550
The total interest paid on the mortgage is
449550 - 270000 = $179550
