V=hpir^2
SA=2pir^2+2hpir
a.
SA=2pi9^2+2*36pi9
SA=2pi81+72pi9
SA=162pi+648pi
SA=810pi in^2
b.
aprox pi=3.14
810*3.14=2542.4cm^2
c.
V=36pi9^2
V=36pi81
V=2916pi in^3
d. 3.14=pi
2916*3.14=9156.24 in^3
<span>Lets say the 1st die rolled a 2 -
there would be 2 combinations for which the sum of dice being < 5 :
2,1
2,2
Now say the 2nd die rolled a 2 -
there would be 2 combinations for which the sum of dice being < 5 :
1,2
2,2
Now we want to count all cases where either dice showed a 2 and sum of the dice was < 5. However note above that the roll (2,2) is counted twice.
So there are three unique dice roll combinations which answer the criteria of at least one die showing 2, and sum of dice < 5:
1,2
2,1
2,2
The total number of unique outcomes for two dice is 6*6=36 .
So, the probability you are looking for is 3/36 = 1/12</span>
Answer:
-3+x=x+4
Step-by-step explanation:
Pls Mark Brainliest
Answer:
A. 12+2p
G. p+p+12
Step-by-step explanation:
The Eagles basketball team scored 12 more than 2 times as many points in the last game of the season than in the first game
Number of points scored in the last game of the season = 2p + 12
A. 12+2p
Equivalent
B. 2+p+12
= P + 14
Not equivalent
C. 2+12p
Not equivalent
D. p(2+12)
= 2p + 12p
Not equivalent
E. p+2+12
= P + 14
Not equivalent
F. 12p+2p
Not equivalent
G. p+p+12
= 2p + 12
Equivalent
|-1/5| < |-1/3| < |-7|
Your answer:
B. y = -1/5 x^2; y = -1/3 x^2; y = -7x^2
Look at the picture.
