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3 years ago

5

Subtract the second equation from the first 6x + 5y =16 - (6x + 2y=10)

1 answer:

Akimi4 [234]3 years ago

5 0

Answer:

3x = 6

Step-by-step explanation:

Combine:

6x + 5y =16

- (6x + 2y=10)

--------------------

3y = 6, so that y = 2 (We weren't actually asked to find the solution.)

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On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)

xz_007 [3.2K]

Answer:

$P = 10 + 2\sqrt{53}$ units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3), X (2, 3),

Y (4, -4), Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

For WX:

$(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)$

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

$WX = \sqrt{(-5)^2 + (0)^2}$

$WX = \sqrt{25}$

$WX = 5$

For XY:

$(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)$

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

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

$XY = \sqrt{-2^2 + 7^2}$

$XY = \sqrt{4 + 49}$

$XY = \sqrt{53}$

For YZ:

$(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)$

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

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

$YZ = \sqrt{7^2 + (-2)^2}$

$YZ = \sqrt{49 + 4}$

$YZ = \sqrt{53}$

For ZW:

$(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)$

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

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

$ZW = \sqrt{0^2 + (-5)^2}$

$ZW = \sqrt{0 + 25}$

$ZW = \sqrt{25}$

$ZW = 5$

The Perimeter (P) is as follows:

$P = WX + XY + YZ + ZW$

$P = 5 + \sqrt{53} + \sqrt{53} + 5$

$P = 5 + 5 + \sqrt{53} + \sqrt{53}$

$P = 10 + 2\sqrt{53}$ units

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3 years ago

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Points A, B, and C are collinear. Find the value of x given AB = 3x, BC = 5x + 10. and AC = 74.

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Answer:

x = 8

Step-by-step explanation:

since it's a collinerar

or, AB+BC=AC

or, 3x+5x+10 = 74

or, 8x = 74-10

or, 8x = 64

or, x = 64/8

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Answer:

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3 years ago

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