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Furkat [3]
4 years ago
9

If cos (A) = 1/2 with A in QIV, find sec (A/2).

Mathematics
1 answer:
Ludmilka [50]4 years ago
6 0
If A is QIV, then        3π/2 ≤A≤2π; 
we have to find out in what quadrant is A/2
          (3π/2)/2≤A/2≤(2π)/2 ⇒ 3π/4≤A/2≤π
We can see that A/2 will be in QIII; therefore the sec (A/2) will be negative (-).

1) we have to calculate cos (A/2)
Cos (A/2)=⁺₋√[(1+cos A/2)/2]
We choose this formula: Cos (A/2)= -√[(1+cos A/2)/2],  because sec A/2 is in quadrant Q III, and the secant (sec A/2=1/cos A/2) in this quadrant is negative.

Cos (A/2)=-√[(1+cos A)/2]=-√[(1+(1/2)]/2=-√(3/4)=-(√3)/2.

2) we compute the sec (A/2)
Data:
cos (A/2)=-(√3)/2

sec (A/2)=1/cos (A/2)
sec (A/2)=1/(-(√3)/2)=-2/√3=-(2√3)/3

Answer: sec (A/2)=-(2√3)/3
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Does anyone know the answer to this ??? I got the first part of it, help with the second part pls !!!
Ad libitum [116K]

Solving the quadratic function, it is found that the particle returns to the ground after 7 seconds.

<h3>What is the quadratic function for the particle's height?</h3>

The particle's height after t seconds is modeled by the following equation:

s(t) = -16t² + v(0)t.

In which v(0) is the initial velocity of the particle, which in this problem is of 112 ft/s, hence:

s(t) = -16t² + 112t.

The particle hits the ground when s(t) = 0, hence:

s(t) = 0

-16t² + 112t = 0

-16t(t - 7) = 0.

Hence the non-trivial solution is:

t - 7 = 0 -> t = 7.

The particle returns to the ground after 7 seconds.

More can be learned about quadratic functions at brainly.com/question/24737967

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5 0
2 years ago
Are the terms 1a and 1b like terms?
Anettt [7]
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3 years ago
What is the length of line ED?
oee [108]

Answer:

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Step-by-step explanation:

2x+4 = (3/2)x+6

2x = (3/2)x+2

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8 0
3 years ago
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Oksana_A [137]

Answer:

the answer is 6 i believe

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5 0
4 years ago
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