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bonufazy [111]
1 year ago
9

3) Use the graph to determine if the associated function would have a POSITIVEor NEGATIVE leading coefficient.N-1-3O positiveneg

ativeO cannot be determined
Mathematics
1 answer:
defon1 year ago
7 0

it is an exponential function, of the form:

\begin{gathered} f(x)=ax^k+bx^{k-1}+\cdots+ck^{k-n} \\ \text{where:} \\ k=2m,m\in N \end{gathered}

since it opens upward:

a>0

Therefore, the coefficient is positive

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4 years ago
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>
<span>= sin(18) = (√5 - 1)/4] </span>
<span>And we seimplify it and it goes: (1/512)*(5-1) = 1/128</span>
4 0
4 years ago
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