QuestionGraph a right triangle with hypotenuse points $(9,-9)$ and $(6,-5)$ . Find the distance between them to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The two points given are $(9,-9)$ and $(6,-5)$ . These two points form the hypotenuse of the right triangle3. We are to find the distance between these two points, which is the length of the hypotenuse

STEP 2

The distance between two points $(x1, y1)$ and $(x2, y2)$ in a plane can be calculated using the distance formula derived from the Pythagorean theorem.
$Distance = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

STEP 3

Now, plug in the given values for the points into the distance formula.
$Distance = \sqrt{(6 -9)^2 + (-5 - (-9))^2}$

STEP 4

implify the expressions inside the square root.
$Distance = \sqrt{(-3)^2 + (4)^2}$

STEP 5

Square the values inside the square root.
$Distance = \sqrt{9 +16}$

STEP 6

Add the values inside the square root.
$Distance = \sqrt{25}$

STEP 7

The square root of25 is5, so the distance between the two points is5 units.
$Distance =5$ The distance between the points $(9,-9)$ and $(6,-5)$ is5 units.

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QuestionGraph a right triangle with hypotenuse points $(9,-9)$ and $(6,-5)$ . Find the distance between them to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The two points given are $(9,-9)$ and $(6,-5)$ . These two points form the hypotenuse of the right triangle3. We are to find the distance between these two points, which is the length of the hypotenuse

STEP 2

The distance between two points $(x1, y1)$ and $(x2, y2)$ in a plane can be calculated using the distance formula derived from the Pythagorean theorem.
$Distance = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

STEP 3

Now, plug in the given values for the points into the distance formula.
$Distance = \sqrt{(6 -9)^2 + (-5 - (-9))^2}$

STEP 4

implify the expressions inside the square root.
$Distance = \sqrt{(-3)^2 + (4)^2}$

STEP 5

Square the values inside the square root.
$Distance = \sqrt{9 +16}$

STEP 6

Add the values inside the square root.
$Distance = \sqrt{25}$

STEP 7

The square root of25 is5, so the distance between the two points is5 units.
$Distance =5$ The distance between the points $(9,-9)$ and $(6,-5)$ is5 units.

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QuestionGraph a right triangle with hypotenuse points (9,−9)(9,-9)(9,−9) and (6,−5)(6,-5)(6,−5). Find the distance between them to the nearest tenth.

Studdy Solution

STEP 1

Assumptions1. The two points given are (9,−9)(9,-9)(9,−9) and (6,−5)(6,-5)(6,−5). These two points form the hypotenuse of the right triangle3. We are to find the distance between these two points, which is the length of the hypotenuse

STEP 2

The distance between two points (x1,y1)(x1, y1)(x1,y1) and (x2,y2)(x2, y2)(x2,y2) in a plane can be calculated using the distance formula derived from the Pythagorean theorem.Distance=(x2−x1)2+(y2−y1)2Distance = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}Distance=(x2−x1)2+(y2−y1)2​

STEP 3

Now, plug in the given values for the points into the distance formula.Distance=(6−9)2+(−5−(−9))2Distance = \sqrt{(6 -9)^2 + (-5 - (-9))^2}Distance=(6−9)2+(−5−(−9))2​

STEP 4

implify the expressions inside the square root.Distance=(−3)2+(4)2Distance = \sqrt{(-3)^2 + (4)^2}Distance=(−3)2+(4)2​

STEP 5

Square the values inside the square root.Distance=9+16Distance = \sqrt{9 +16}Distance=9+16​

STEP 6

Add the values inside the square root.Distance=25Distance = \sqrt{25}Distance=25​

STEP 7

The square root of25 is5, so the distance between the two points is5 units.Distance=5Distance =5Distance=5The distance between the points (9,−9)(9,-9)(9,−9) and (6,−5)(6,-5)(6,−5) is5 units.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionGraph a right triangle with hypotenuse points $(9,-9)$ and $(6,-5)$ . Find the distance between them to the nearest tenth.

Assumptions1. The two points given are $(9,-9)$ and $(6,-5)$ . These two points form the hypotenuse of the right triangle3. We are to find the distance between these two points, which is the length of the hypotenuse

The distance between two points $(x1, y1)$ and $(x2, y2)$ in a plane can be calculated using the distance formula derived from the Pythagorean theorem.
$Distance = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

Now, plug in the given values for the points into the distance formula.
$Distance = \sqrt{(6 -9)^2 + (-5 - (-9))^2}$

implify the expressions inside the square root.
$Distance = \sqrt{(-3)^2 + (4)^2}$

Square the values inside the square root.
$Distance = \sqrt{9 +16}$

Add the values inside the square root.
$Distance = \sqrt{25}$

The square root of25 is5, so the distance between the two points is5 units.
$Distance =5$ The distance between the points $(9,-9)$ and $(6,-5)$ is5 units.