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3 years ago

(12,-1),(-2,-3) What is the finding slope?

Mathematics

2 answers:

ddd [48]3 years ago

7 0

Answer:

The slope is 1/7 or m=1/7

Step-by-step explanation:

-3-(-1)/-2-12

-3+1/-2-12

-2/-14

1/7

viktelen [127]3 years ago

5 0

Answer: 7

Step-by-step explanation:

m = -14 / -2 = 7

You might be interested in

14x-5=7+11x i need help with this

leva [86]

Answer:

x=4

Step-by-step explanation:

first lets move it over

3x = 12

x=4

check:

14*4 = 56

-5 =

7+ 11x=

11*4 = 44

44+7 = 51

51 = 51

hope this helps :D

3 0

3 years ago

Plz help! Estimate. Then find the sum or difference.

ANTONII [103]

=743.5−462.87
=743.5+−462.87
=280.63

98.01+52.003
=150.013

74.9−16.227
=74.9+−16.227
=58.673

4 0

3 years ago

What are the coordinates of the vertices of the pre-image given? ry= −x ◦ T1, −2(x, y)

nataly862011 [7]

<h3><u>Answer:</u></h3>

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

<h3><u>Step-by-step explanation:</u></h3>

We have to find the coordinates of the vertices of the pre-image given?

Ry= −x ◦ T_1, −2(x, y)

i.e. we have to find the composition of reflection along the line y=-x and translation with the rule:

(x,y) → (x+1,y-2)

Now we have coordinates of A",B",C" and D" as:

A"(-4,-2)

B"(-2,-1)

C"(-2,-4)

D"(-4,-3)

Now we are asked to find the coordinates of the pre-image.

Also when any point is reflected along y=-x then the point is transformed to:

(x,y) → (-y,-x)

Let A,B,C and D are the points of the pre-image.

Hence, the coordinates of the pre-image are given as:

so, the transformation is given as:

A→A'→A"

B→B'→B"

C→C'→C"

D→D'→D"

Where A',B',C',D' represents the transformation after translation and A",B",C",D" represents the transformation after reflection as well.

The coordinates of A' are (2,4)

B' are (1,2)

C' are (4,2)

and D' are (3,4)

Now, the coordinates of pre-image are given as:

A'(x,y) → A(x-1,y+2)=A(1,6)

B'(x,y) → B(x-1,y+2)=B(0,4)

C'(x,y) → C(x-1,y+2)=C(3,4)

and D'(x,y) → D(x-1,y+2)=D(2,6)

Hence, the coordinates of pre-image are:

A(1,6) , B(0,4) , C(3,4) , D(2,6)

4 0

3 years ago

The center of a hyperbola is (−4,3) , and one vertex is (−4,7) . The slope of one of the asymptotes is 2.

Monica [59]

Answer:

The answer to your question is below

Step-by-step explanation:

C (-4, 3)

V (-4, 7)

asymptotes = 2 = $\frac{b}{a}$

- This is a vertical hyperbola, the equation is

$\frac{(y - k)^{2} }{a^{2} } + \frac{(x - h)^{2} }{b^{2} } = 1$

slope = 2

a is the distance from the center to the vertex = 4

b = 2(4) = 8

$\frac{(y - 3)^{2} }{4^{2} } + \frac{(x + 4)^{2} }{8^{2} } = 1$

$\frac{(y - 3)^{2} }{16} + \frac{(x + 4)^{2} }{64} = 1$

7 0

3 years ago

Read 2 more answers

Please help !

makvit [3.9K]

Answer:

New line =3*ZX

Step-by-step explanation:

It is given that the isosceles trapezoid is dilated using a scale factor of 1/3

and a line is drawn through the center of the new dilated figure, so

let us suppose that line is EF which is drawn and it will also pass through

the center in similar way the line ZX passes through so after considering this situation, we can come to this point

$\text{new Line EF}=\frac{ZX}{1/3} \\EF=3*ZX$

so new line will be 3 times longer than ZX.

3 0

3 years ago