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stich3 [128]
3 years ago
13

What is the nth term in 3,9,15,21,27?

Mathematics
1 answer:
PtichkaEL [24]3 years ago
6 0
Look at the equation backwards, its minus 6 every time. so therefore the ninth term would be 51. double check yourself count up + 6 til you get to the ninth term! :)
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What is the sum of the following numbers. -7,5,-3,6,-4,7,-5,3-1,4​
Sav [38]

Answer:

Step-by-step explanation:

Tổng : -7,5+(-3,6)+(-4,7)+(-5,3)+(-1,4)=-22,5

3 0
3 years ago
Solve 3 + 4x > -5 for x
Debora [2.8K]

you have the following inequality:

3 + 4x > -5

in order to solve for x, proceed as follow:

3 + 4x > -5 subtract 3 both sides

4x > - 5 - 3

4x > -8 divide by 4 both sides

x > -8/4

x > -2

Hence, the solution to the given inequality is x > -2

7 0
1 year ago
Hey can u help me how to do this plz.
Svetach [21]
Okay so x represents the amount of tickets sold. So first, you’d add the $40 + $70 which equals $110. So x = $110.

Now we have to solve for y. Same thing but with different numbers. $200 + $260 = $460.
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x. y.
4 0
3 years ago
You coach a basketball ball team of 12 players; 5 players must be on the floor at all times; Figuring that every player can play
makvit [3.9K]
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.
4 0
3 years ago
F(x)=1/2×-2, what is f(14)
tamaranim1 [39]
F(14) would be, 14=1/2x-2
7 0
3 years ago
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