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3 years ago

Find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7)

Mathematics

2 answers:

sergeinik [125]3 years ago

8 0

Answer:

Hope this helps :)

Step-by-step explanation:

Nitella [24]3 years ago

5 0

The equation of a hyperbola is:

(x – h)^2 / a^2 - (y – k)^2 / b^2 = 1

So what we have to do is to look for the values of the variables:

For the given hyperbola : center (h, k) = (0, 0)
a = 3(distance from center to vertices)
a^2 = 9

c = 7 (distance from center to vertices; given from the foci)
c^2 = 49

By the hypotenuse formula:
c^2 = a^2 + b^2
b^2 = c^2 - a^2

b^2 = 49 – 9

b^2 = 40

Therefore the equation of the hyperbola is:

(x^2 / 9) – (y^2 / 40) = 1

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Answer:

10 ft 10 in

Step-by-step explanation:

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3 years ago

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Answer:

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A grocery store clerk put only cans of tomato soup and cans of chicken soup on a shelf. The ratio of the number of cans of chick

larisa [96]

Answer:

9 cans of tomato soup

Step-by-step explanation:

We are told that only cans of tomato soup and cans of chicken soup are put on the shelf.

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In fraction, this is 8/17

Now, this implies that there are 8 cans of chicken noodle soup while there are a total of 17 cans on the shelf.

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3 years ago

The set of ordered pairs represents a

Murrr4er [49]

Answer:

The rule that represents the function is $y=x^2+1$ therefore the function is $f(x)=x^2+1$

Step-by-step explanation:

We have 5 ordered pairs in the plane xy. This means that every pair has the form (x, y).

Then, we have 5 values of x, which will give us 5 values of y, using the rule that represents the function.

The easy evaluation is that when x=0, the value of y is y=1, and then we can evaluate the rule for x=-1, and x=1, the value of y is the same, y=2. We can see here that we have a parabolic function, that is not centered in the origin of coordinates because when x=0, y=1.

So we propose the rule $y=x^2+1$ which is correct for the first 3 values of x.

Now, we evaluate the proposed rule when x=2, and when x=3. This evaluations can be written as

$f(2)=2^2+1=5$