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PilotLPTM [1.2K]
3 years ago
14

HELP TIMER Write the equation of a hyperbola centered at the origin with x-intercept +/- 4 and foci of +/-2(squareroot 5)

Mathematics
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:

\frac{x2}{a} - \frac{y2}{b2} = 1

Step-by-step explanation:

A hyperbola is the locus of a point such that its distance from a point to two points (known as foci) is a positive constant.

The standard equation of a hyperbola centered at the origin with transverse on the x axis is given as:

\frac{X2}{16} - \frac{b}{4} = 1

The coordinates of the foci is at (±c, 0), where c² = a² + b²

Given that  a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2√5. Since the x intercept is ±4, this means that at y = 0, x = 4. Substituting in the standard equation:

I don't feel like explaining so...

a. = 4

The foci c is at +/-2√5, using c² = a² + b²:

B = 2

Substituting the value of a and b to get the equation of the hyperbola:

\frac{x2}{a2} -      \frac{y2}{b2} = 1  

\frac{x2}{16} - \frac{b2}{4} = 1

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What’s the correct answer for this?
otez555 [7]

Answer:

0.7 + 0.4 - 0.2 = 0.9

Step-by-step explanation:

Let's denote the probabilities as following:

The probability that the show had animals is

P(A) = 0.7

The probability that the show aired more than 10 times is

P(B) = 0.4

The probability that the show had animals and aired more than 10 times is

P(A⋂B) = 0.2

The probability that a randomly selected show had animals or aired more than 10 times is P(A⋃B)

The correct form of  addition rule to determine the probability that a randomly selected show had animals or aired more than 10 times is:

P(A⋃B) = P(A) + P(B) - P(A⋂B) = 0.7 + 0.4 - 0.2 = 0.9

=> Option B is correct

Hope this helps!

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3 years ago
How would I go about solving this?
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Answer:

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6 0
3 years ago
pam is making fruit punch for a party using the ratio of 2 cups of club soda with 5 cups of juice. complete the table to show eq
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Answer:

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Step-by-step explanation:

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3 years ago
Consider the vector b⃗ b→b_vec with length 4.00 mm at an angle 23.5∘∘ north of east. What is the y component bybyb_y of this vec
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Answer:

  • \large\boxed {1.59 mm}

Explanation:

<u>1. Given vector:</u>

  • length: 4.00 mm = magnitude of the vector
  • angle: 23.5º north of east = 23.5º from the x-axys (counterclockwise)

<u>2. y-component</u>

The y-component may be determined using the sine ratio, the angle from the x-axys (counterclockwise direction), and the magnitude of the vector.

  • sine (23.5º) = y-component / magnitude

  • y-component = magnitude × sine (23.5º) = 4.00 mm × sine (23.5º) = 1.59 mm.

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Expanded form.2(b + c) *
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Answer:

Step-by-step explanation:

2b + 2c

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