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Olin [163]
3 years ago
13

Anna is no more than 3 years older than 2 times Jamie’s age. Jamie is at least 14 and Anna is at most 35. Which system of linear

inequalities can be used to find the possible ages of Anna, a, and Jamie, j?
a ≥ 3 + 2j; j ≥ 14, a ≤ 35
a ≤ 3 + 2j; j ≥ 14, a ≤ 35
a ≥ 3 + 2j; j ≤ 14, a ≤ 35
a ≤ 3 + 2j; j ≤ 14, a ≤ 35
Mathematics
2 answers:
joja [24]3 years ago
8 0

Answer: The answer is (b) a ≤ 3 + 2j; j ≥ 14, a ≤ 35.

Step-by-step explanation:  Given that Anna is no more than 3 years older than 2 times Jamie’s age. Jamie is at least 14 and Anna is at most 35. We are to select the correct combination of inequalities among the given options.

Also, 'a' and 'j' are the possible ages of Anna and Jamie respectively. Therefore, according to the given information, we can write

a\leq 2j+3,\\j\geq 14,\\a\leq 35.

Thus, the correct option is (b) a ≤ 3 + 2j; j ≥ 14, a ≤ 35.

Butoxors [25]3 years ago
6 0

Answer:

b is correct on e d g e n u i t y

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Evaluate : <br> \(\cos 12 \cos 24 \cos 36 \cos 48 \cos 72 \cos 84\)
NeTakaya
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>
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