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3 years ago

8

Two-variable linear equations 2x – 5y = 20

1 answer:

Lynna [10]3 years ago

7 0

Y= (2x-20)/ 5 then you replace y so with a simple calculation u get x than u replace x with it value to get y as simple as that

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3 years ago

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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

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3 years ago

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