Mathematics, 23.01.2022 01:00 viktoria1198zz

$\boxed{\boxed{ \begin{array}{}\large\tt The \\ \tt \large Distance \: Formula \: \\ \large \tt and \\ \large\tt \: Coordinate \: Proof \end{array}}}$ Direction: Find the distance between each pair of points on the coordinate plane and write the solution.

1. A(3, 8) and B(3, -2)

2. O(5, -2) and K(7, -2)

3. L(4, -4) and M(-4, 6)

4. P(5, 1) and C(2, 3)

5. Q(-10, 9) and R(-2, 3)

$\boxed{\boxed{ \begin{array}{}\large\tt Nonsense + Useless \\ \large \tt= Report! \end{array}}}$

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Answer from: zumar

Distance formula

$\boxed{\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Now

$\\ \tt\Rrightarrow AB=\sqrt{(3-3)^2+(-2-8)^2}=\sqrt{100}=10units$

$\\ \tt\Rrightarrow OK=\sqrt{(5-7)^2+(-2+2)^2}=\sqrt{4}=2units$

$\\ \tt\Rrightarrow LM=\sqrt{(-4-4)^2+(6+4)^2}=\sqrt{64+100}=\sqrt{164}\approx 13units$

$\\ \tt\Rrightarrow PC=\sqrt{(5-2)^2+(1-2)^2}=\sqrt{9+1}=\sqrt{10}=3.1units$

$\\ \tt\Rrightarrow QR=\sqrt{(-2+10)^2+(3-9)^2}=\sqrt{64+36}=\sqrt{100}=10units$

Answer from: BluSeaa

Answer of all given questions are given below :

1) AB = 10 units 2) OK = 2 units3) LM = 2√41 units4) PC ≈ 3.6 units5) QR = 10 units

Step-by-step explanation:

Here's the required formula to find distance between points :

$\star{\small{\underline{\boxed{\sf{\red{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}}}$

According to this formula, we'll solve all the given questions and find the distance between points.

1. A(3, 8) and B(3, -2)

Substituting all the given values in the formula to find the distance between points:

${\implies{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\implies{\small{\sf{AB = \sqrt{\Big(3 - 3 \Big)^{2} + \Big( - 2 - 8 \Big)^{2}}}}}}$

${\implies{\small{\sf{AB = \sqrt{\Big(0\Big)^{2} + \Big( - 10 \Big)^{2}}}}}}$

${\implies{\small{\sf{AB = \sqrt{\Big(0 \times 0\Big) + \Big( - 10 \times - 10 \Big)}}}}}$

${\implies{\small{\sf{AB = \sqrt{\big(0 + 100\big)}}}}}$

${\implies{\small{\sf{AB = \sqrt{100}}}}}$

${\implies{\sf{\underline{\underline{\purple{AB = 10}}}}}}$

Hence, the distance between points AB is 10 units..

$\begin{gathered}\end{gathered}$

2. O(5, -2) and K(7, -2)

Substituting all the given values in the formula to find the distance between points:

${\longrightarrow{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\Big(7 - 5\Big)^{2} + \Big( - 2 + 2 \Big)^{2}}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\Big(2\Big)^{2} + \Big(0 \Big)^{2}}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\Big(2 \times 2\Big) + \Big(0 \times 0 \Big)}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{\big(4 + 0 \big)}}}}}$

${\longrightarrow{\small{\sf{OK = \sqrt{4}}}}}$

${\longrightarrow{\sf{\underline{\underline{\pink{OK = 2}}}}}}$

Hence, the distance between points OK is 2 units.

$\begin{gathered}\end{gathered}$

3. L(4, -4) and M(-4, 6)

Substituting all the given values in the formula to find the distance between points:

${\longmapsto{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\Big( - 4 - 4\Big)^{2} + \Big( 6 + 4 \Big)^{2}}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\Big( - 8\Big)^{2} + \Big(10 \Big)^{2}}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\Big( - 8 \times - 8\Big) + \Big(10 \times 10\Big)}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{\big( 64 + 100\big)}}}}}$

${\longmapsto{\small{\sf{LM = \sqrt{164}}}}}$

${\longmapsto{\sf{\underline{\underline{\orange{LM = 2 \sqrt{41} }}}}}}$

Hence, the distance between points LM is 2√41 units.

$\begin{gathered}\end{gathered}$

4. P(5, 1) and C(2, 3)

Substituting all the given values in the formula to find the distance between points:

${\dashrightarrow{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\Big(2 - 5 \Big)^{2} + \Big(3 - 1 \Big)^{2}}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\Big( - 3\Big)^{2} + \Big(2\Big)^{2}}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\Big( - 3 \times - 3\Big) + \Big(2 \times 2\Big)}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{\big( 9 + 4\big)}}}}}$

${\dashrightarrow{\small{\sf{PC = \sqrt{13}}}}}$

${\dashrightarrow{\sf{\underline{\underline{\green{PC \approx 3.6}}}}}}$

Hence, the distance between points PC is 3.6 units.

$\begin{gathered}\end{gathered}$

5. Q(-10, 9) and R(-2, 3)

Substituting all the given values in the formula to find the distance between points:

${\twoheadrightarrow{\small{\sf{Distance = \sqrt{\Big(x_{2} - x_{1} \Big)^{2} + \Big(y_{2} - y_{1} \Big)^{2}}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\Big( - 2 + 10 \Big)^{2} + \Big(3 - 9 \Big)^{2}}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\Big( 8 \Big)^{2} + \Big( - 6\Big)^{2}}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\Big( 8 \times 8 \Big)+ \Big( - 6 \times - 6\Big)}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{\big(64 + 36\big)}}}}}$

${\twoheadrightarrow{\small{\sf{QR = \sqrt{100}}}}}$

${\twoheadrightarrow{\sf{\underline{\underline{\blue{QR = 10}}}}}}$

Hence, the distance between points QR is 10 units.

$\rule{300}{2.5}$

Answer from: Quest

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.

Answer from: Quest

If i think this is what your asking, he should look at the grams or pounds/ounces and compare the to their cost then max can identify which one is proportional and which is not ( i think xd)

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