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klio [65]
4 years ago
13

Two figures are similar. The ratio of the corresponding sides between the figure is 9 to 12. Based on the ratio, which could be

the scale factor used to dilate one figure to create the other?
F. 0.3
G. 3.4
H. 3
J. 0.75
Mathematics
1 answer:
topjm [15]4 years ago
7 0

Answer:

The answer is J

Step-by-step explanation:

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What point on the line y=9x+8 is closest to the origin
Marat540 [252]

to find the point, find the location

realize that the shortest line that crosses through the origin and y=9x+8 is perpendicular to y=9x+8

perpendicular lines have slopes that multiply to get -1

y=9x+8 has a slope of 9

9*m=-1, m=-1/9

the slope of the mystery line is -1/9

since it passes through the origin, the y intercept is 0

y=(-1/9)x is the equation of the line from the point to the origin

find the intersection of this line and the original line

set them equal to each other

(-1/9)x=9x+8

multily both sides by 9

-x=81x+72

minus 81 both sides

-82x=72

divide both sides by -82

x=-36/41

find y

y=(-1/9)x

y=(-1/9)(-36/41)

y=4/41

the point is (\frac{-36}{41},\frac{4}{41})

3 0
3 years ago
Determine the number of real solutions for each of the given equations. Equation Number of Solutions y = -3x2 + x + 12 y = 2x2 -
rosijanka [135]

Answer:

Step-by-step explanation:

Our equations are

y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\

Let us understand the term Discriminant of a quadratic equation and its properties

Discriminant is denoted by  D and its formula is

D=b^2-4ac\\

Where

a= the coefficient of the x^{2}

b= the coefficient of x

c = constant term

Properties of D: If D

i) D=0 , One real root

ii) D>0 , Two real roots

iii) D<0 , no real root

Hence in the given quadratic equations , we will find the values of D Discriminant  and evaluate our answer accordingly .

Let us start with

y = -3x^2 + x + 12\\a=-3 , b =1 , c =12\\D=1^2-4*(-3)*(12)\\D=1+144\\D=145\\D>0 \\

Hence we have two real roots for this equation.

y = 2x^2 - 6x + 5\\

y = 2x^2 - 6x + 5\\a=2,b=-6,c=5\\D=(-6)^2-4*2*5\\D=36-40\\D=-4\\D

Hence we do not have any real root for this quadratic

y = x^2 + 7x - 11\\a=1,b=7,-11\\D=7^2-4*1*(-11)\\D=49+44\\D=93\\

Hence D>0 and thus we have two real roots for this equation.

y = -x^2 - 8x - 16\\a=-1,b=-8,c=-16\\D=(-8)^2-4*(-1)*(-16)\\D=64-64\\D=0\\

Hence we have one real root to this quadratic equation.

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The answer is in the attachment

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Temka [501]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

here's the solution in attachment.

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