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tino4ka555 [31]
3 years ago
6

Determine the graph foci and asymptote equation of x^2/4-y^2=1

Mathematics
2 answers:
larisa86 [58]3 years ago
8 0

Answer:

B)

Foci: ((sqrt5),0),((-sqrt5),0)

Asymptotes: y=((1/2)x), y=((-1/2)x)

Step-by-step explanation:

Same rule with x before y as I mentioned in the other problem to identify the graph.

JulsSmile [24]3 years ago
7 0

Answer:

A

Step-by-step explanation:

this is the answer

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Step-by-step explanation:

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3 years ago
What is the difference in length between a 1 1/4 inch button and a 3/8 inch button?
charle [14.2K]
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3 years ago
Jon's closet has 3 black shirts, 8 blue shirts, 6 black pants and 7 blue pants. He wants to determine the probability of selecti
algol [13]

Answer:

Jon did determine the probability correctly.

Step-by-step explanation:

<em>In Jon's closet, there are a total of 24 clothes. Among those pieces of clothing, there are a total of 9 out of 24 clothes that are black. There are also 11 shirts out of 24 clothes. </em>

<em />

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<em>Therefore, Jon is correct.</em>

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3 0
3 years ago
#54 Simplify and write the answers using positive exponents only.
kramer
Gg easy

remember
(x^m)/(x^n)=x^(m-n)
and
(x^m)^n=x^(mn)
and
(xyz)^m=(x^m)(y^m)(z^m)
and
(m/n)^a=(m^a)/(n^a)
and
x^-n=1/(x^n)

so
( \frac{6mn^{-2}}{3m^{-1}n^2} )^{-3}=
( \frac{6}{3} )^{-3}( \frac{m}{m^{-1}} )^{-3}( \frac{n^{-2}}{n^2} )^{-3}=
(2^{-3})((m^2)^{-3})((n^{-4})^{-3})=
( \frac{1}{2^3})(m^{-6})(n^{12})=
( \frac{1}{8} )( \frac{1}{m^6})(n^{12})=
\frac{n^{12}}{8m^6}

8 0
3 years ago
Identify all sequences of transformations that results in similar figures.
Rus_ich [418]
A is the answer. Reflection and rotation.
7 0
2 years ago
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