To find the answer, plot down the factors for every number.
12: 1, 2 ,3 ,4, 6, 12
18: 1, 2, 3, 6, 9, 18
84: 1, 2, 3, 4, 6, 7, 12
If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF
Answer:
Object 2, which has a density of 1.9 g/cm3, since it has more density than freshwater.
The mass would be the same
47kg on the moon as well
The harbour contains salt water while the river contains
fresh water. So assuming that the densities of fresh water and salt water are:
density (salt water) = 1029 kg / m^3
density (fresh water) = 1000 kg / m^3
The amount of water (in mass) displaced by the barge
should be equal in two waters.
mass displaced (salt water) = mass displaced (fresh
water)
Since mass is also the product of density and volume, therefore:
<span>[density * volume]_salt water = [density * volume]_fresh
water ---> 1</span>
First we calculate the amount of volume displaced in the harbour
(salt water):
V = 3.0 m * 20.0 m * 0.70 m
V = 42 m^3 of salt water
Plugging in the values into equation 1:
1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water
Volume fresh water displaced = 43.218 m^3
Therefore the depth of the barge in the river is:
43.218 m^3 = 3.0 m * 20.0 m * h
<span>h = 0.72 m (ANSWER)</span>
We are given with the x and y components of Vector A and B. In this case, we compute the resultant of both components of each vector. The vector is equal to the square root of the sum of the squares of the components. A is equal to 2.60 cm. B is equal to 5.56 cm. B is found in quadrant Iv and has an angle of 42.447 degrees as a terminal angle. A has an angle of 59.98 degrees.
a. 5.6082 < -15.53 degreesc. 6.63 <-64.98 degreesb. x = 6.63 cos -64.98 degrees = 2.80 y = 6.63 sin -64.98 degrees = -6.00
