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deff fn [24]
3 years ago
14

K\3 +21=27 how do I solve this equation? can someone help me

Mathematics
1 answer:
Natalija [7]3 years ago
6 0
K/3 +21 = 27
subtract 21 from both sides...
k/3 = 48
now multiply 48 and 3 to find k...
48 x 3 = 144
k=144
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Write an inequality to represent the phrase: The sum of a number an 6 is no less than 31
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The inequality that represents this phrase is n+6 > 31.
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I need help with this plz
emmasim [6.3K]

Answer:

x^{15}

Step-by-step explanation:

Distribute the ^3 on top

(27 x^{6} y^{-9})

distribute the ^ -9 on bottom, only to what is in the parenthesis

(27 x^{-9} y^{-9})

the 27's cross out, the y^{-9} 's cross out

x^{6}/ x^{-9} is left

move the  x^{-9} to the top of fraction to change its sign

x^{6} · x^{9} = x^{15}

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Find three consecutive integers such that three times the first added to the last is 22
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Consetive integers are 1 apart
they are
x,x+1,x+2

3 times the first added to the last is 22
3x+x+2=22
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4x+2=22
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4 0
3 years ago
In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?
kupik [55]

Answer:

There are 5,586,853,480 different ways to select the jury.

Step-by-step explanation:

The order is not important.

For example, if we had sets of 2 elements

Tremaine and Tre'davious would be the same set as Tre'davious and Tremaine. So we use the combinations formula.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In how many different ways can a jury of 12 people be randomly selected from a group of 40 people?

Here we have n = 40, x = 12.

So

C_{40,12} = \frac{40!}{12!(28)!} = 5,586,853,480

There are 5,586,853,480 different ways to select the jury.

3 0
3 years ago
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