Let x = greater number
y = smaller number
(1)
(1)
(2)
We'll substitute y in (1) to (2)
(2)
x - 9 = 0 or x + 3 = 0
x = 9 x = -3
and
y = x - 6 y = x - 6
y = 9 - 6 y = -3 - 6
y = 3 y = -9
Therefore, the two numbers can be 9 and 3 or -3 and -9.
To calculate this probability we must take into account that there is the same probability that any of the 3 urns is chosen.
This probability is:
P (U1) = P (U2) = P (U3) = 1/3
Urn 1 contains 7 black and 3 red marbles
Urn 2 contains 2 black and 8 marbles network
Urn 3 contains 5 black marbles and 5 red marbles.
The probability of obtaining a black marble in Urn 1 is 7/10.
The probability of obtaining a black marble in Urn 2 is 2/10
The probability of obtaining a black marble in Urn 3 is 5/10.
Then we look for the probability of obtaining a black marble from urn 1 or a black marble from urn 2 or a black marble from urn 3. This is:
P (U1yB) + P (U2yB) + P (U3yB)
So:
(1/3) * (7/10) + (1/3) * (2/10) + (1/3) * (5/10) = 0,2333 + 0,0667 + 0,1667 = 0, 4667.
The probability that it is a black marble is 46.67%
8) A. x + 2,
2) B. f(x) = (x -3) (x +9).
Step-by-step explanation:
Step 1; If the function equals 0, we substitute all the separate factors to 0. We keep the variable on the left side while all the other values go to the right side.
If f(x) = (x +3) (x -9) = 0, then x +3 = 0, x -9 = 0. So x = -3 and x = 9.
If f(x) = (x -3) (x +9) = 0, then x -3 = 0, x +9 = 0. So x = 3 and x = -9.
If f(x) = (x -3) (x -9) = 0, then x -3 = 0, x -9 = 0. So x = 3 and x = 9.
If f(x) = (x +3) (x +9) = 0, then x +3 = 0, x +9 = 0. So x = -3 and x = -9.
So if the factors are 3 and -9, option B is an answer.
Step 2; For the given graph, the same applies. The graph passes through the points of (-2, 0) and (4, 0) on the x-axis.
So if x = -2, then x +2 = 0 and if x = 4, x -4 = 0. The first equation is option A while the second equation is not given.
0.8333 , 1 mile
but try to make sure of it ^
Answer:
25+3x simplified
Step-by-step explanation:
HOPE THIS HELPS YOU OUT! BE THE FIRST TO COMMENT /FREE ON THESE COMMENTS TO GET A FREE BRAINLIY AWARD! PLZ ALSO CONDISER MARKING ME BRAINLIEST AND THANKS! MY JOURNEY HAS JUST BEGAN ON BRAINLY SO ASK ME IF U HAVE ANY QUESTIONS! -BEACHVIBES23
