The space between the two spheres will be the volume of the larger sphere minus the volume of the smaller sphere. Given that the volume of any sphere is:
V=(4πr^3)/3 The space between to sphere of different radius and positioned about the same center is:
S=(4πR^3)/3-(4πr^3)/3 I used S=volume of space, R=larger radius and r=smaller radius...
S=(4π/3)(R^3-r^3), we are told that R=5 and r=4 so
S=(4π/3)(5^3-4^3)
S=(4π/3)(125-64)
S=(4π/3)(61)
S=244π/61
S=4π cm^3
S≈12.57 cm^3 (to nearest hundredth of a ml)
- 18
- 6
- 2
- 3
- 4
- 1
- 5
- 3
Step-by-step explanation:
<em>it</em><em>'s</em><em> </em><em>a</em><em> </em><em>m</em><em>atter</em><em> </em><em>of</em><em> </em><em>substi</em><em>tuting</em><em> </em><em>the</em><em> </em><em>val</em><em>ues</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>,</em><em>b</em><em> </em><em>,</em><em> </em><em>c</em><em> </em><em>into</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>8</em><em> </em><em>expressions</em>
Answer:
Step-by-step explanation:
Expand
Apply log properties to expand the natural log
When the log terms are in multiplication then we break the terms and add it
use ln for each term
Apply the above property to break the terms
Now move the exponent before ln
The first box has a total of 10 items ( 3 pencils + 7 pens = 10)
The probability of picking a pen would be 7/10
The second box has a total of 8 items ( 4 pencils + 4 crayons = 8)
The probability of picking a crayon would be 4/18, which reduces to 1/2
To find the probability of doing both, multiply each probability together:
7/10 x 1/2 = 7/20
The probability is 7/20
148/x=100/108(148/x)*x=(100/108)*x -148=0.925925925926*x
(0.925925925926) to get x148/0.925925925926=x 159.84=x x=159.84
