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kenny6666 [7]
3 years ago
11

Evaluate : \(\cos 12 \cos 24 \cos 36 \cos 48 \cos 72 \cos 84\)

Mathematics
1 answer:
NeTakaya3 years ago
4 0
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>
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If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes
Keith_Richards [23]
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
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4 years ago
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Lily bought 25.29 pounds of grapefruit. The lightest grapefruit weighed 1.4 pounds. The heaviest grapefruit weighed 1.6 pounds.
Gemiola [76]

Answer:

B

Step-by-step explanation:

GIven that the lightest is 1.4 and the heaviest is 1.6, this implies that ALL the fruit must weigh between 1.4 and 1.6.

This also implies that the average weight of the fruit must also be between 1.4 lb and 1.6 lb.

recall that : average weight = total weight / number of fruit

we are given than total weight = 25.29 lb, so

average weight = 25.29 / number of fruit

Just go down the choices and test which choice gives an average weight between 1.4 and 1.6 lbs.

Choice A: number of fruit = 19

average = 25.29 / 19 = 1.33  (outside of range, not the answer)

Choice B: number of fruit = 17

average = 25.29 / 17 =1.49   (within range , possible answer)

Choice C: number of fruit = 10

average = 25.29 / 10 = 2.529 (outside of range, not the answer)

Choice D: number of fruit = 50

average = 25.29 / 50 = 0.5058  (outside of range, not the answer)

From the above, we can see that only B gives an average between 1.4 and 1.6 lb.

5 0
3 years ago
-12x + 12 = 4 - 15x + 9
emmainna [20.7K]

Answer:

x = 1/3

Step-by-step explanation:

-12x + 12 = 4 - 15x + 9

Combine like terms

-12x +12 = -15x+13

Add 15x to each side

-12x+12 +15x = -15x+15x+13

3x+12 = 13

Subtract 12 from each side

3x+12-12 = 13-12

3x = 1

Divide by 3

3x/3 = 1/3

x = 1/3

8 0
3 years ago
Which of the following points lies on the circle whose center is at the origin and whose radius is 10?
Elodia [21]

Unfortunately, you haven't shared "the following points."


The equation of a circle centered at the origin and whose radius is 10 is


x^2 + y^2 = 10^2.

8 0
3 years ago
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