P( 8,2) & Q( 3,8)
Apply Pythagoras.
d^2 = (8 - 3)^2 + ( 2 - 8)^2
d^2 = 5^2 + (-6)^2
d^2 = 25 + 36
d^2 = 61
d = sqrt(61) ~ 7.81 (2 d.p.
Using the distance formula it is the square root of 61 which is about 7.8 to the nearest tenth
7.8
7,8
Its radius is 4.8 cm, rounded to the nearest tenth.
A circle has a circumference of 110.6 mm. Find its diameter to the nearest tenth.
tan-1(0.8) = 38.65980825 degrees or 38.7 degrees to the nearest tenth.
100
The surface area, to the nearest tenth, of a sphere that has a diameter of 14 inches is: 615.8 inches2
The distance between the points of (2, 3) and (7, 0) is the square root of 34
Using the distance formula it is the square root of 61 which is about 7.8 to the nearest tenth
caca
61
A. 29 b. 6.7 c. 25 d. 5.4
6.7
The distance is 6.4 units.
It is 4.2
i think 4
square root of (64 + 289) = 18.8 to the nearest tenth of a unit
29.0
What is 1.4 to the nearest tenth