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Afina-wow [57]
3 years ago
10

I need to find the distance between (3,4) , (7,2)

Mathematics
1 answer:
fiasKO [112]3 years ago
4 0

Answer:

2\sqrt{5}

Step-by-step explanation:

Use the distance formula.

Subtract the x-coordinates and square the difference.

3 - 7 = -4; (-4)^2 = 16

Subtract the y-coordinates and square the difference.

4 - 2 = 2; 2^2 = 4

Add the differences and take the square root of the sum.

16 + 4 = 20; sqrt(20) = sqrt(4 * 5) = 2sqrt(5)

distance = 2sqrt(5)

You can use the distance formula which does the same thing.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

d = \sqrt{(7 - 3)^2 + (2 - 4)^2}

d = \sqrt{(4)^2 + (-2)^2}

d = \sqrt{16 + 4}

d = \sqrt{20}

d = \sqrt{4 \times 5}

d = \sqrt{4} \times \sqrt{5}

d = 2\sqrt{5}

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Nat2105 [25]

Answer:

250

Step-by-step explanation:

1. 190 students is the number that is 100-24 = 76% of the 6th grade population

0.76x = 190

x = (190/.76) = <u>250</u>

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3 years ago
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Amber’s bird feeder holds 4/5 of a cup
Harlamova29_29 [7]

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0.8 units

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4 years ago
Calculate the average (arithmetic mean) of the list of numbers shown in the table. (rounded to the nearest tenth) A) 35.2 B) 39.
jenyasd209 [6]
<h3><u>Given Numbers:-</u></h3>
  • 23, 33 ,33 ,27 ,25 ,68 ,27 ,44 ,72

<h3><u>To Find:-</u></h3>
  • average of this set of numbers

<h3><u>Solution</u>:-</h3>

\green{ \underline { \boxed{ \sf{Average = \frac{Sum \:of \: observation}{Number \:of \: observation}}}}}

\longrightarrow Here, no of observation are 9

Now,

\begin{gathered}\implies \small \sf Average =\frac{23+33+33+27+25+68+27+44+72}{9}\\\end{gathered}

\begin{gathered}\\\implies \small \sf Average =\frac{56+60+93+71+72}{9}\\\end{gathered}

\begin{gathered}\\\implies \sf Average =\frac{116+164+72}{9}\\\end{gathered}

\begin{gathered}\\\implies \sf Average =\frac{352}{9}\\\end{gathered}

\begin{gathered}\\\implies \sf Average = 39.1 \\\end{gathered}

\longrightarrowThe Average of given set of numbers is 39.1 .

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3 0
2 years ago
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Find y as a function of x if y"' + 9y' = 0, y(0) = 6, y'(0) = 9, y"(0) = 18. y(x) =
Andreas93 [3]

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y(0) = 6

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<u>6 = C₁ + C₂ .......................................................1</u>

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<u>y' = -3C₂sin 3x + 3C₃cos 3x</u>

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<u>C₁ = 8</u>

Thus,

<u>y(x) = 8 - 2cos 3x + 3sin 3x</u>

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USPshnik [31]
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