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NemiM [27]
3 years ago
8

3. Jack buys a bicycle on sale for $59.

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

Answer:

One 50$, one 5$, and four 1$ bills

marin [14]3 years ago
3 0

Answer:

$50, $5, $2, $2

Step-by-step explanation:

50+5=55

2+2=4

55+4=59

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Y varies directly as x . when x=3 , then y=12 . find y when x=20.
Phoenix [80]

Answer:

Congress. would equal 80 due to x being worth 1/4 of y's total.

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2 years ago
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barxatty [35]

Answer:

A table?

Step-by-step explanation:

fingers crossed this is right

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3 years ago
Which image shows the correct position of M(-4, 3)?
pickupchik [31]

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Graph A

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is the equation of the line, in slope-intercept form, that passes through (4, 2) and (-2, -3)?
Vlad1618 [11]

Answer:

y = 5/6x -4/3

Step-by-step explanation:

We can find the slope

m = (y2-y1)/(x2-x1)

    = (-3-2)/(-2-4)

    =-5/-6

    = 5/6

We can use point slope form to find the equation

y-y1 = m(x-x1)

y-2 = 5/6(x-4)

Distribute

y-2 = 5/6x - 10/3

Add 2 to each side

y-2+2 = 5/6x -10/3+2

Get a common denominator

y = 5/6x -10/3 +6/3

y = 5/6x -4/3

6 0
3 years ago
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

Learn more about hyperbola here

brainly.com/question/7098764

#SPJ4

4 0
1 year ago
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