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3 years ago

Simplify 2m - [n - (m - 2n)].

1 answer:

7 0

Let's simplify step-by-step.

2m−(n−(m−2n))

Distribute the Negative Sign:

=2m+−1(n−(m−2n))

=2m+−1n+−1(−m)+−1(2n)

=2m+−n+m+−2n

Combine Like Terms:

=2m+−n+m+−2n

=(2m+m)+(−n+−2n)

=3m+−3n

Answer:

=3m−3n

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One number is six more than the other number . thesum of their squares is 90

Andreyy89

Answer:

the answer is 3 and 9. 9 is 6 greater than 3 and 9 squared and 3 squared added up give you 90.

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1 year ago

Write an equation for the line perpendicular to y=1/3x-2 and passing through the point (3,3)

maria [59]

line given slope = x coefficient = 1/3

Perpendicular slopes must be opposite reciprocals of each other: m1 * m2 = –1

line slope (m) = -3

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from the point given

x = 3

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2 years ago

Please help, I need this quickly. Fairly easy, help pleasee

ahrayia [7]

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3 years ago

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of

notsponge [240]

Answer:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

A(0, 0), B(2, 0) and C(0, 2).

What would be the coordinates of triangle A'B'C' if triangle ABC was dilated by a factor of 1/3 ?

=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, $(x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)$

So, The coordinates of the triangle A'B'C' are:

$A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})$

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3 years ago

Find the perimeter of WXYZ. Round to the nearest tenth if necessary.

yanalaym [24]

Answer:

C. 15.6

Step-by-step explanation:

Perimeter of WXYZ = WX + XY + YZ + ZW

Use the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to calculate the length of each segment.

✔️Distance between W(-1, 1) and X(1, 2):

Let,

$W(-1, 1) = (x_1, y_1)$

$X(1, 2) = (x_2, y_2)$

Plug in the values

$WX = \sqrt{(1 - (-1))^2 + (2 - 1)^2}$

$WX = \sqrt{(2)^2 + (1)^2}$

$WX = \sqrt{4 + 1}$

$WX = \sqrt{5}$

$WX = 2.24$

✔️Distance between X(1, 2) and Y(2, -4)

Let,

$X(1, 2) = (x_1, y_1)$

$Y(2, -4) = (x_2, y_2)$

Plug in the values

$XY = \sqrt{(2 - 1)^2 + (-4 - 2)^2}$

$XY = \sqrt{(1)^2 + (-6)^2}$

$XY = \sqrt{1 + 36}$

$XY = \sqrt{37}$

$XY = 6.08$

✔️Distance between Y(2, -4) and Z(-2, -1)

Let,

$Y(2, -4) = (x_1, y_1)$

$Z(-2, -1) = (x_2, y_2)$

Plug in the values

$YZ = \sqrt{(-2 - 2)^2 + (-1 -(-4))^2}$

$YZ = \sqrt{(-4)^2 + (3)^2}$

$YZ = \sqrt{16 + 9}$

$YZ = \sqrt{25}$

$YZ = 5$

✔️Distance between Z(-2, -1) and W(-1, 1)

Let,

$Z(-2, -1) = (x_1, y_1)$

$W(-1, 1) = (x_2, y_2)$

Plug in the values

$ZW = \sqrt{(-1 -(-2))^2 + (1 - (-1))^2}$

$ZW = \sqrt{(1)^2 + (2)^2}$

$ZW = \sqrt{1 + 4}$

$ZW = \sqrt{5}$

$ZW = 2.24$

✅Perimeter = 2.24 + 6.08 + 5 + 2.24 = 15.56

≈ 15.6

5 0

3 years ago