Math / Geometry

QuestionFind the distance between the pair of points $(0,-\sqrt{6})$ and $(\sqrt{14}, 0)$ . If necessary, express the answer in simplified radical form and then round to two decimal places.
The distance between the given points is $\square$ units. (Simplify your answer. Type an exact answer, using radicals as needed.)

Studdy Solution

STEP 1

What is this asking? We need to find the distance between two points on a coordinate plane and give the answer in both simplified radical form and as a decimal rounded to two places. Watch out! Don't forget to simplify the radical and round correctly!
Also, remember the distance is always positive.

STEP 2

1. Apply the distance formula
2. Simplify the radical
3. Calculate the decimal approximation

STEP 3

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

STEP 4

Let's call our first point $(x_1, y_1) = (0, -\sqrt{6})$ and our second point $(x_2, y_2) = (\sqrt{14}, 0)$ .
It doesn't matter which point is which, as long as we're consistent!

STEP 5

Now, let's plug in our values: $d = \sqrt{(\sqrt{14} - 0)^2 + (0 - (-\sqrt{6}))^2}.$

STEP 6

See how those double negatives become a plus?
Awesome! Let's keep simplifying: $d = \sqrt{(\sqrt{14})^2 + (\sqrt{6})^2}.$ Squaring a square root removes the radical, so we get: $d = \sqrt{14 + 6}.$ Adding those together gives us: $d = \sqrt{20}.$

STEP 7

We can simplify $\sqrt{20}$ by finding the prime factorization of $20$ .
We have $20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$ .

STEP 8

So, $\sqrt{20} = \sqrt{2^2 \cdot 5} = \sqrt{2^2} \cdot \sqrt{5} = 2\sqrt{5}$ .
We've pulled out the perfect square!

STEP 9

Now, let's calculate the decimal approximation of $2\sqrt{5}$ .
Using a calculator, we find that $\sqrt{5} \approx 2.236$ .

STEP 10

Multiplying by $2$ , we get $2 \cdot 2.236 = 4.472$ .
Rounding to two decimal places, we get 4.47.

STEP 11

The distance between the points is $2\sqrt{5}$ units, which is approximately 4.47 units.

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Studdy Solution

STEP 1

What is this asking? We need to find the distance between two points on a coordinate plane and give the answer in both simplified radical form and as a decimal rounded to two places. Watch out! Don't forget to simplify the radical and round correctly!
Also, remember the distance is always positive.

STEP 2

1. Apply the distance formula
2. Simplify the radical
3. Calculate the decimal approximation

STEP 3

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

STEP 4

Let's call our first point $(x_1, y_1) = (0, -\sqrt{6})$ and our second point $(x_2, y_2) = (\sqrt{14}, 0)$ .
It doesn't matter which point is which, as long as we're consistent!

STEP 5

Now, let's plug in our values: $d = \sqrt{(\sqrt{14} - 0)^2 + (0 - (-\sqrt{6}))^2}.$

STEP 6

See how those double negatives become a plus?
Awesome! Let's keep simplifying: $d = \sqrt{(\sqrt{14})^2 + (\sqrt{6})^2}.$ Squaring a square root removes the radical, so we get: $d = \sqrt{14 + 6}.$ Adding those together gives us: $d = \sqrt{20}.$

STEP 7

We can simplify $\sqrt{20}$ by finding the prime factorization of $20$ .
We have $20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$ .

STEP 8

So, $\sqrt{20} = \sqrt{2^2 \cdot 5} = \sqrt{2^2} \cdot \sqrt{5} = 2\sqrt{5}$ .
We've pulled out the perfect square!

STEP 9

Now, let's calculate the decimal approximation of $2\sqrt{5}$ .
Using a calculator, we find that $\sqrt{5} \approx 2.236$ .

STEP 10

Multiplying by $2$ , we get $2 \cdot 2.236 = 4.472$ .
Rounding to two decimal places, we get 4.47.

STEP 11

The distance between the points is $2\sqrt{5}$ units, which is approximately 4.47 units.

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Studdy Solution

STEP 1

STEP 2

1. Apply the distance formula2. Simplify the radical3. Calculate the decimal approximation

STEP 3

The distance ddd between two points (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​) is given by the **distance formula**: d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.d=(x2​−x1​)2+(y2​−y1​)2​.

STEP 4

Let's call our first point (x1,y1)=(0,−6)(x_1, y_1) = (0, -\sqrt{6})(x1​,y1​)=(0,−6​) and our second point (x2,y2)=(14,0)(x_2, y_2) = (\sqrt{14}, 0)(x2​,y2​)=(14​,0).It doesn't matter which point is which, as long as we're consistent!

STEP 5

Now, let's **plug in** our values: d=(14−0)2+(0−(−6))2.d = \sqrt{(\sqrt{14} - 0)^2 + (0 - (-\sqrt{6}))^2}.d=(14​−0)2+(0−(−6​))2​.

STEP 6

STEP 7

We can **simplify** 20\sqrt{20}20​ by finding the **prime factorization** of 202020.We have 20=2⋅2⋅5=22⋅520 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 520=2⋅2⋅5=22⋅5.

STEP 8

So, 20=22⋅5=22⋅5=25\sqrt{20} = \sqrt{2^2 \cdot 5} = \sqrt{2^2} \cdot \sqrt{5} = 2\sqrt{5}20​=22⋅5​=22​⋅5​=25​.We've pulled out the perfect square!

STEP 9

Now, let's **calculate** the decimal approximation of 252\sqrt{5}25​.Using a calculator, we find that 5≈2.236\sqrt{5} \approx 2.2365​≈2.236.

STEP 10

Multiplying by 222, we get 2⋅2.236=4.4722 \cdot 2.236 = 4.4722⋅2.236=4.472.Rounding to two decimal places, we get **4.47**.

STEP 11

The distance between the points is 252\sqrt{5}25​ units, which is approximately **4.47** units.

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1. Apply the distance formula
2. Simplify the radical
3. Calculate the decimal approximation

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

Let's call our first point $(x_1, y_1) = (0, -\sqrt{6})$ and our second point $(x_2, y_2) = (\sqrt{14}, 0)$ .
It doesn't matter which point is which, as long as we're consistent!

Now, let's plug in our values: $d = \sqrt{(\sqrt{14} - 0)^2 + (0 - (-\sqrt{6}))^2}.$

We can simplify $\sqrt{20}$ by finding the prime factorization of $20$ .
We have $20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$ .

So, $\sqrt{20} = \sqrt{2^2 \cdot 5} = \sqrt{2^2} \cdot \sqrt{5} = 2\sqrt{5}$ .
We've pulled out the perfect square!

Now, let's calculate the decimal approximation of $2\sqrt{5}$ .
Using a calculator, we find that $\sqrt{5} \approx 2.236$ .

Multiplying by $2$ , we get $2 \cdot 2.236 = 4.472$ .
Rounding to two decimal places, we get 4.47.

The distance between the points is $2\sqrt{5}$ units, which is approximately 4.47 units.