QuestionGraph a right triangle using points $(-6,-3)$ and $(-1,9)$ . Find the distance between these points to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The two given points (-6,-3) and (-1,9) form the hypotenuse of the right triangle. . The distance between the two points is the length of the hypotenuse.
3. The distance is calculated using the Euclidean distance formula.

STEP 2

The Euclidean distance formula in two dimensions is given by $Distance = \sqrt{(x2-x1)^2 + (y2-y1)^2}$ where $(x1,y1)$ and $(x2,y2)$ are the coordinates of the two points.

STEP 3

Substitute the coordinates of the given points into the formula.
$Distance = \sqrt{(-1-(-6))^2 + (9-(-3))^2}$

STEP 4

implify the expressions inside the brackets.
$Distance = \sqrt{()^2 + (12)^2}$

STEP 5

Square the numbers inside the square root.
$Distance = \sqrt{25 +144}$

STEP 6

Add the numbers inside the square root.
$Distance = \sqrt{169}$

STEP 7

Take the square root of169.
$Distance =13$ The distance between the points (-6,-3) and (-1,9) is13 units.

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QuestionGraph a right triangle using points $(-6,-3)$ and $(-1,9)$ . Find the distance between these points to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The two given points (-6,-3) and (-1,9) form the hypotenuse of the right triangle. . The distance between the two points is the length of the hypotenuse.
3. The distance is calculated using the Euclidean distance formula.

STEP 2

The Euclidean distance formula in two dimensions is given by $Distance = \sqrt{(x2-x1)^2 + (y2-y1)^2}$ where $(x1,y1)$ and $(x2,y2)$ are the coordinates of the two points.

STEP 3

Substitute the coordinates of the given points into the formula.
$Distance = \sqrt{(-1-(-6))^2 + (9-(-3))^2}$

STEP 4

implify the expressions inside the brackets.
$Distance = \sqrt{()^2 + (12)^2}$

STEP 5

Square the numbers inside the square root.
$Distance = \sqrt{25 +144}$

STEP 6

Add the numbers inside the square root.
$Distance = \sqrt{169}$

STEP 7

Take the square root of169.
$Distance =13$ The distance between the points (-6,-3) and (-1,9) is13 units.

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QuestionGraph a right triangle using points (−6,−3)(-6,-3)(−6,−3) and (−1,9)(-1,9)(−1,9). Find the distance between these points to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The two given points (-6,-3) and (-1,9) form the hypotenuse of the right triangle. . The distance between the two points is the length of the hypotenuse.3. The distance is calculated using the Euclidean distance formula.

STEP 2

The Euclidean distance formula in two dimensions is given byDistance=(x2−x1)2+(y2−y1)2Distance = \sqrt{(x2-x1)^2 + (y2-y1)^2}Distance=(x2−x1)2+(y2−y1)2​where (x1,y1)(x1,y1)(x1,y1) and (x2,y2)(x2,y2)(x2,y2) are the coordinates of the two points.

STEP 3

Substitute the coordinates of the given points into the formula.Distance=(−1−(−6))2+(9−(−3))2Distance = \sqrt{(-1-(-6))^2 + (9-(-3))^2}Distance=(−1−(−6))2+(9−(−3))2​

STEP 4

implify the expressions inside the brackets.Distance=()2+(12)2Distance = \sqrt{()^2 + (12)^2}Distance=()2+(12)2​

STEP 5

Square the numbers inside the square root.Distance=25+144Distance = \sqrt{25 +144}Distance=25+144​

STEP 6

Add the numbers inside the square root.Distance=169Distance = \sqrt{169}Distance=169​

STEP 7

Take the square root of169.Distance=13Distance =13Distance=13The distance between the points (-6,-3) and (-1,9) is13 units.

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QuestionGraph a right triangle using points $(-6,-3)$ and $(-1,9)$ . Find the distance between these points to the nearest tenth.

Assumptions1. The two given points (-6,-3) and (-1,9) form the hypotenuse of the right triangle. . The distance between the two points is the length of the hypotenuse.
3. The distance is calculated using the Euclidean distance formula.

The Euclidean distance formula in two dimensions is given by $Distance = \sqrt{(x2-x1)^2 + (y2-y1)^2}$ where $(x1,y1)$ and $(x2,y2)$ are the coordinates of the two points.

Substitute the coordinates of the given points into the formula.
$Distance = \sqrt{(-1-(-6))^2 + (9-(-3))^2}$

implify the expressions inside the brackets.
$Distance = \sqrt{()^2 + (12)^2}$

Square the numbers inside the square root.
$Distance = \sqrt{25 +144}$

Add the numbers inside the square root.
$Distance = \sqrt{169}$

Take the square root of169.
$Distance =13$ The distance between the points (-6,-3) and (-1,9) is13 units.