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Sunny_sXe [5.5K]
2 years ago
8

0.3, 0.08, 0.003, 0.83 In order from smallest to largest

Mathematics
2 answers:
pishuonlain [190]2 years ago
3 0
0.003, 0.08, 0.3, 0.83
Inessa05 [86]2 years ago
3 0

Answer:

0.003, 0.08, 0.3, 0.83

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Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0
Xelga [282]
Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
\left(h+c,\:k\right),\:\left(h-c,\:k\right)

\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

9x^2-y^2-36x-4y+23=0 \ \textgreater \  \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}
9x^2-36x-4y-y^2=-23

\mathrm{Factor\:out\:coefficient\:of\:square\:terms}
9\left(x^2-4x\right)-\left(y^2+4y\right)=-23

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1
\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \  \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \  Refine
\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1

Now rewrite in hyperbola standardform
\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1

\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3
\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)

Now we must compute c
\sqrt{1^2+3^2} \ \textgreater \  \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \  1^2 = 1 \ \textgreater \  \sqrt{1+3^2}

3^2 = 9 \ \textgreater \  \sqrt{1+9} \ \textgreater \  \sqrt{10}

Therefore the hyperbola foci is at \left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)

For the vertices we have \left(2+1,\:-2\right),\:\left(2-1,\:-2\right)

Simply refine it
\left(3,\:-2\right),\:\left(1,\:-2\right)
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0
3 years ago
Winston's team participated in a total of 22 chess matches. Winston
Zepler [3.9K]

Answer:

Peter Jonathan Winston (March 18, 1958 – disappeared January 26, 1978) was an American chess player from New York City

Step-by-step explanation:

In late 1977, Winston attended a FIDE-rated tournament at Hunter College High School in New York City. Despite being one of the highest-rated players in the tournament, Winston lost all nine of his games. A few months later, on January 26, 1978, following further surprising game losses, Peter Winston vanished and was never heard from again. According to some sources, Winston's disappearance occurred when he left his home without money, identification, or luggage during a severe winter storm. Many chess players who were close to or acquainted with Winston claim that the champion chess player's mental health had deteriorated, along with his game performance, in the last few years of his life, and that the decline in his mental health may have led to his disappearance.

3 0
2 years ago
Read 2 more answers
How many sixteenths sure in 15/16
Feliz [49]

Answer:

15

Step-by-step explanation:

there are 15 sixteenths in 15/16, how you figure this out is the numerator tells you how many are in the fraction, in this case the numerator says 15, so there are 15, sixteenths in, 15/16

8 0
3 years ago
An isosceles triangle has (5 points) A, at least two congruent sides B. two sides that are perpendicular C. three sides with the
aev [14]

Answer:

A. at least two congruent sides

Step-by-step explanation:

Since, we know that,

'An isosceles triangle is a triangle having two sides of equal length or at least two sides of equal length (special case is equilateral triangle)'.

Thus, from the options, we see that,

Options B, C and D are not correct.

<em>Since, congruent sides means that the lengths of the sides are equal.</em>

Thus, we get that,

'An isosceles triangle has at least two congruent sides'.

Hence, option A is correct.

4 0
3 years ago
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Jack is 4 times as old as lacey. 7 years ago the sum of their age was 61. how old are they now?
BaLLatris [955]
I am guessing Jack is 51 and Lacey is 17.

7 0
2 years ago
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