Answer:
4000 cm^3
Step-by-step explanation:
Note that no image was attached for reference purposes
Step one:
We are given that the original length and width of the cube is 20cm
if the cut part is 10cm
hence the dimension of the cut part is
lenght = 20cm
Width= 20cm
Hieght= 10cm
Step two:
the volume of the cut part will be
Volume= 20*20*10
Volume= 4000 cm^3
Answer: 81.85 km/hr
Converting m/sec into km/hr, 25 m/sec = 90 km/hr
Average =
Time = 50/90 = 5/9 hr
next 120 km at an average speed of 80km/h
=> Time = 120/80 = 3/2 hr
last part of its journey at an average speed of 90km/h in 35 Minutes
time = 35 minutes = 35/60 hr = 7/12 hr
Distance covered = 90 * 7 /12 = 105/2 km
Total Distance = 50 + 120 + 105/2 = 435/2 km
Total time = 5/9 + 3/2 + 7/12 = ( 20 + 54 + 21) /36 = 95/36
Average speed = (435/2)/(95 /36) = 432 * 18 /95 = 81.85 km/hr
Answer:
y+30
Step-by-step explanation:
the sum of the numbers is 15y
the new sum after increase each number by 30 is 15y+450
the new mean 15y+450/15
= y+30
We can factor this by using grouping. Take the leading coefficient and multiply it by the constant. In this case we get 5*-7 = -35.
Now we need 2 numbers that add to 2 and multiply to be -35. The numbers are -5 and 7.
So split the 2r into these two terms and group.
(5r^2 - 5r) + (7r - 7)
Factor both groups.
5r(r-1) + 7(r-1)
The factors of (r-1) can be added together to get the answer.
(r-1)(5r+7)
<span>As the age of the U-235 sample is 2.631 billion years, and the half-life of U-235 is 713 million years, the sample has undergone 2.361 X 1,000,000,000 / 713 X 1,000,000 = 3.69 half lives. In each half-life the sample reduces to half its original weight according to the radioactive Half-Life Formula:
ln (Nt /N0) = -kt, where N0 = mass of the original weight of radioactive material, Nt = mass of radioactive material at time t, k = decay constant and t = time interval . We have to put Nt/N0 = 1/2 for time interval = half-life.</span>
