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Artemon [7]
3 years ago
12

What is the distance between he points (5,14) and (18,3)

Mathematics
2 answers:
Gemiola [76]3 years ago
8 0

the distance bewteen points (x1,y1) and (x2,y2) is

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

given (5,14) and (18,3)

d=\sqrt{(18-5)^2+(3-14)^2}

d=\sqrt{(13)^2+(-11)^2}

d=\sqrt{169+121}

d=\sqrt{290}

the distance between those points is √290

Elodia [21]3 years ago
7 0

Answer:

17.03

Step-by-step explanation:

First, find the distance between the x coordinates (13) and the distance between the y coordinates (11). Next, plug these numbers into the Pythagorean theorem (a^2+b^2=c^2) and solve. 13 squared is 169 and 11 squared is 121. Add those together to get 290. The square root of 290 is about 17.02938637 (that's how many digits my calculator can show), and it has no simplified radical form.

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What’s 50 times 5000000000000000000000
kodGreya [7K]

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250000000000000000000000

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Please help me <br>i really want to know the answer<br>and the explanation for this problems​
puteri [66]

Answer:

M = 4.33885225095

Step-by-step explanation:

Area of the square ABFE = 10² = 100

M = 100 - (2P + Q)

Let’s calculate 2P + Q :

The area 2P + Q = area ΔABC + area of sector ACE + area of sector BCF

Note :

ΔABC is an equilateral triangle

m∠CBF = m∠CAE = 30°

area ΔABC  = (CG × AB)÷2 = (8.660254037844×10)÷2 = 43.30127018922

CG = √(10^2 - 5^2)=8.660254037844  (Pythagorean theorem)

area of sector BCF = area ΔACE = 100π ÷ 12 = (8.333333333333)π

then

Area 2P + Q = area  ΔABC + area sector ACE + area sector BCF

                      = 43.30127018922+(100÷12)π+(100÷12)π

                      = 43.30127018922+ (8.333333333333)π + (8.333333333333)π

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M = 100 - (2P + Q) = 100-95.66114774905 = 4.33885225095

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