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alexandr1967 [171]
3 years ago
8

2+2 lalalaoaoaoasooaosa

Mathematics
2 answers:
Tomtit [17]3 years ago
6 0

Answer:

Step-by-step explanation:

4

anyanavicka [17]3 years ago
4 0

Answer:

4

Step-by-step explanation:

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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
Consider the numbers 3/10 and 4/13
serg [7]
Make bototm number same
3/10 times 13/13=39/130
4/13 times 10/10=40/130

A.
39/130 and 40/130
they are 1/130 apart


B.
a ratioal number between
39/130=390/1300
40/130=400/1300
between 390/1300 and 400/1300
any number from 391/1300 to 399/1300 example 395/1300


3 0
4 years ago
jogger ran 2 miles, increase your speed by 1 mph, and then ran another 3 miles. if the total jogging time was one 1/half hours,
notka56 [123]

Answer:

around 5 miles per hour

Step-by-step explanation:

5 0
2 years ago
HELPPPPPPPPP 55 POINTS Isabel graphed the following system of equations.
Arturiano [62]

Answer:

1.  Substitute y = -3x +4 for y in the first equation

2.  Combine like terms and isolate x

3.  Substitute in x into the 2nd equation to get y

Step-by-step explanation:

1.  Substitute y = -3x +4 for y in the first equation

2x- (-3x+4) = 6

2.  Combine like terms

5x -4 = 6

5x -4 +4 = 6+4

5x=10  Divide by 5 to isolate x

5x/5 =10/5

x=2

3. Substitute in x into the 2nd equation to get y

y = -3x +4

y = -3(2) +4

y = -6+4

y=-2

(2,-2)

5 0
3 years ago
The flaplike lateral wall of each atrium is called the
pantera1 [17]
You mean like the tricuspid and mitral (bicuspid) valves?
6 0
3 years ago
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