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Alekssandra [29.7K]

4 years ago

15

Find all points having an x-coordinate of 2 whose distance from the point (-1,-2) is 5

1 answer:

xxTIMURxx [149]4 years ago

3 0

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\
5=\sqrt{(-1-2)^2+(-2-y_1)^2}\\
\sqrt{9+4+4y_1+y_1^2}=5\\
y_1^2+4y_1+13=25\\
y_1^2+4y_1-12=0\\
y_1^2+6y_1-2y_1-12=0\\
y_1(y_1+6)-2(y_1+6)=0\\
(y_1-2)(y_1+6)=0\\
y_1=2 \vee y_1=-6\\\\
\boxed{(2,2),(-6,2)}

$

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3 years ago

Richard walks every day for exercise at a rate of 1 kilometer every 20 minutes.

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3 years ago

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

6 0

4 years ago

Solve the equation<br>4(y-3)=13

I am Lyosha [343]

Step-by-step explanation:

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7 0

3 years ago

Read 2 more answers