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ZanzabumX [31]
3 years ago
13

Please help.. I'm really confused tbh

Mathematics
2 answers:
zzz [600]3 years ago
5 0

X axis is the horizontal axis and y is the vertical.

On a graph, quandrant 1 is the upper right side which requires both x and y to be positive.

Quadrant 2 is upper left, which is negative x and positive y.

Quandrant 3 is lower left, which is negative x and y.

Quadrant 4 is lower right, which is positive x and negative y.

X< 0 is negative and y > 0 is positive.

This matches quadrant 2

The answer is B.

alisha [4.7K]3 years ago
4 0
I think ur answer will be C
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I think the answer is 51.

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Help 20 points!
Sergio [31]

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Step-by-step explanation:

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write the equation in standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0).
Mashcka [7]

Standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is \frac{x^{2} }{144}-\frac{y^{2} }{25}=1

Hyperbola is a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.

Given,

The Vertices of the hyperbola =  (-12,0) and (12,0)

Foci = (-13,0) and (13,0)

a=12

ac=13

c=\frac{13}{12}

We know,

c=\sqrt{1+\frac{b^{2} }{a^{2} } }

c^{2}=1+\frac{b^{2} }{a^{2} }  \\(\frac{13}{12} )^{2}=1+\frac{b^{2} }{144}  \\\frac{169}{144}=1+ \frac{b^{2} }{144} \\\frac{b^{2} }{144} =\frac{25}{144}\\ b^{2}=25

The equation of the hyperbola is

\frac{x^{2} }{a^{2} }-\frac{y^{2} }{b^{2} }=1\\  \frac{x^{2} }{144 }-\frac{y^{2} }{25 }=1

Hence, the standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is \frac{x^{2} }{144}-\frac{y^{2} }{25}=1

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4 0
2 years ago
The perimeter of a rectangle is 56 inches it's with is 6 inches less than its length find the lenth and the width
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P = 2(L + W)
P = 56
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56 = 2(L + L - 6)
56 = 2(2L - 6)
56 = 4L - 12
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W = L - 6
W = 17 - 6
W = 11 <== width (W) = 11 inches

7 0
3 years ago
What is the solution to this system? 2x+3y=12 -4x+5y=-2
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answer: (3,2)

step by step explanation: :)

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3 years ago
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