Please help! Vectors and angles
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See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0
<h3>How to graph the hyperbola?</h3>The equation of the hyperbola is given as:
12x^2 - 3y^2 - 108 = 0
Start by calculating the transverse axis
So, we have:
<u>Transverse axis</u>
The vertices of the given hyperbola are (0, 0)
This means that
(h, k) = 0
Where
a = 3 and b = 6
The transverse axis is calculated as:
y = ±b/a(x - h) + k
So, we have:
y = ±6/3(x - 0) + 0
Evaluate the difference and sum
y = ±6/3x
Evaluate the quotient
y = ±2x
This means that the transverse axes are y = 2x and y =-2x
<u>The vertices</u>
In the above section, we have:
The vertices of the given hyperbola are (0, 0)
This means that
(h, k) = 0
<u>The co-vertices</u>
In the above section, we have:
The vertices of the given hyperbola are (0, 0)
This means that
(h, k) = 0
And
a = 3 and b = 6
The co-vertices are
(h - a, k) and (h + a, k)
So, we have:
(0 - 3, 0) and (0 + 3, 0)
Evaluate
(-3,0) and (3, 0)
See attachment for the graph of the hyperbola
Read more about hyperbola at:
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Every value{0.01, 0.29, 85%} can represent the probability of an event occurring except option d that is 1.5.
Given to us,
a.)
b.) 0.29
c.) 85%
d.)
The probability help us to know about the probability of specific events occurring.
For a sure event, the probability is always 1,
while for an event that will never happen the probability is always 0.
Thus, probability(p), .
Now looking at the options,
a.) = 0.01
b.) 0.29
c.) 85% = 0.85
d.) = 1.5
Now comparing each option with .
Therefore, the only option which is not feasible is option d that is 1.5.
Hence, every value{0.01, 0.29, 85%} can represent the probability of an event occurring except option d that is 1.5.
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