Answer:
<h2>181 440</h2>
Step-by-step explanation:
We have 9 choices for Choosing the first object
8 choices for Choosing the 2nd object
.
.
.
3 choices for Choosing the seventh object
Therefore if we want to choose7 objects without replacement from 9 objects we have : 9×8×7×6×5×4×3 = 181 440
Also ,we can calculate it this way :
9P7 = 181 440
Answer:
- b. y = 2/3x +5
- c. y = -3 or -1/2
Step-by-step explanation:
b. The particualar steps for solving a linear equation such as this may vary from one equation to another. In general, you want to put all the y-terms on one side of the equal sign and everything else on the other side. Here's how we'll do that in this case.
Use the distributive property to eliminate parentheses.
... 6x -3y +12 = 4x -3
Find the unwanted terms on the side of the equation where y is, then add their opposite. In this case, we're adding -(6x+12)
... -3y = (4x -3) -(6x +12) = -2x -15
Divide by the coefficient of y.
... y = -2x/(-3) -15/(-3)
... y = 2/3x +5
____
c. This is a 2nd-degree (quadratic) equation with y as its only variable. The factor (2y+1) appears in both terms so we can use the distributive property to factor it out. Then the equation becomes ...
... (y +3)(2y +1) = 0
The zero product rule tells us this product will be zero only when one or the other of the factors is zero. This fact helps us find the values of y that are the solution to the equation.
For y+3 = 0:
... y + 3 = 0
... y = -3 . . . . . . add -3 to both sides of the equation
For 2y +1 = 0:
... 2y +1 = 0
... y +1/2 = 0 . . . . divide by 2
... y = -1/2 . . . . . . add -1/2
The solutions are y = -3 or y = -1/2.
Part (a)
Label the doors A, B, C.
If a person goes through door A, then they have 2 choices (B or C) to pick as an exit. Similarly, if they go through door B, then they have 2 choices for an exit (A or C). Finally, if they go through door C, then they have 2 choices for an exit (A or B).
We have 3 choices for the entrance, and 2 choices for the exit.
So overall, 3*2 = 6 different combinations are possible.
Those 6 combos are
AB, AC, BA, BC, CA, CB
<h3>Answer: 6</h3>
=============================================
Part (b)
This is effectively the same as part (a), but we can now re-use the entrance door. For any door we enter, we have 3 choices to pick from for the exit. There are 3 doors, so 3 entrances and 3 exits, meaning 3*3 = 9 different combinations.
Those 9 combos are
AA, AB, AC, BA, BB, BC, CA, CB, CC
<h3>Answer: 9</h3>
