Answer:
-176.........................
Answer:
Step-by-step explanation:
Answer:
The first one is equivalent [the 6x+48 = 2(3x+24)] and the second one is <u>NOT</u> equivalent [the 7x+21 ≠ 2(5x+3)]
Step-by-step explanation:
Just follow distributive property to solve these. You can ignore the first expression in both until you have to compare the answers.
1. 6x+48 and 2(3x+24)
2(3x+24) ---> 2(3x) + 2(24) ---> <u>6x + 48</u>
Bring in the first expression ~ <u>6x+48 and 6x+48 </u>
They are the same, so they are equivalent
2. 7x+21 and 2(5x+3)
2(5x+3) ---> 2(5x) + 2(3) ----> 10x + 6
Bring in the first expression ~ <u>7x+21 and 10x + 6</u>
They are NOT the same, so they are NOT equivalent
Answer:
Make a table with values of x and y using the equation y + 3 = -1/4(x - 3) :
x y
0 -2.25
3 -3
5 -3.5
-1 -2
-9 0
Now, graph these and draw a line through the points:
The function "choose k from n", nCk, is defined as
nCk = n!/(k!*(n-k)!) . . . . . where "!" indicates the factorial
a) No position sensitivity.
The number of possibilities is the number of ways you can choose 5 players from a roster of 12.
12C5 = 12*11*10*9*8/(5*4*3*2*1) = 792
You can put 792 different teams on the floor.
b) 1 of 2 centers, 2 of 5 guards, 2 of 5 forwards.
The number of possibilities is the product of the number of ways, for each position, you can choose the required number of players from those capable of playing the position.
(2C1)*(5C2)*(5C2) = 2*10*10 = 200
You can put 200 different teams on the floor.
