Given:
Positions of two artifacts are at points (1, 4) and (5, 2).
To find:
The distance between these two artifacts.
Solution:
Distance formula: The distance between two points is
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
Using distance formula, the distance between two points (1, 4) and (5, 2) is
![d=\sqrt{(5-1)^2+(2-4)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%285-1%29%5E2%2B%282-4%29%5E2%7D)
![d=\sqrt{(4)^2+(-2)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%284%29%5E2%2B%28-2%29%5E2%7D)
![d=\sqrt{16+4}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B16%2B4%7D)
![d=\sqrt{20}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B20%7D)
![d=4.4721359](https://tex.z-dn.net/?f=d%3D4.4721359)
Round to the nearest tenth of a unit.
![d\approx 4.5](https://tex.z-dn.net/?f=d%5Capprox%204.5)
Therefore, the distance between two artifacts is 4.5 units.
Answer:
x=10.5
Step-by-step explanation:
if the two triangles were ratios it would be like
19.3:7.2 to x:3.9
so 75.27 = 7.2x
whcih is 10.45 which is approximately 10.5
The probability of getting all heads is 1 / 2^6 = 1/64 as there is only 1 event where this happens in a possible 2^6 = 64 events. It is the same as the probability of getting all tails. The probability of getting at least 1 head is 1 - p(all tails) = 63/64.
Answer:
v= -12
Step-by-step explanation:
v= -12
v = -29+17
v= -12
810
Step-by-step explanation:
multiple 3x3 9 then multiple 3x9 27 multiply 27x3 equal 81 then add zero 810