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alexandr402 [8]
2 years ago
12

Three consecutive integers have a sum of 30. Which equation can be used to find x, the value of the smallest of the three number

s? (x + 1) + (x + 2) = 30 x + (x + 1) + (x + 2) = 30 x (x + 1) (x + 2) = 30 3 x (x + 1) (x + 2) = 30
Mathematics
1 answer:
castortr0y [4]2 years ago
7 0

Answer:

X+(X+1)+X+2)=30

Step-by-step explanation:

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9 members of a club. Need 3 officers. How many combinations are possible.
irina [24]

Answer:

I think its 3 combos

Step-by-step explanation:

9÷3=3 combos?

Hope this helps :D

3 0
3 years ago
Hayley begins solving this problem by combining like terms. 3x + 5x = 10 Which problem begins the same way? A) 5x = 20 Eliminate
GarryVolchara [31]

Answer:

In a linear equation  of one variable there are two terms terms containing variable and constant.

Variable means term containing x,y,z which act as variables. and constant means any rational number . In higher classes you can include real number also.

The given equation is 3x + 5x = 10

So, the solution out of four options is A) 5x = 20.

Here variable is on one side of equation and constant on other side.

But Option (D) looks absolutely correct if we replace ? by either + sign or Negative (-) sign.




8 0
3 years ago
Read 2 more answers
Which is a reasonable domain for the function f(x)=10x+8
Soloha48 [4]

Answer:

(-∞,∞)

Step-by-step explanation:

The domain is all possible x values of the function

There are no constraints or holes for this and therefore the domain is all real numbers

8 0
3 years ago
Brainiest fastest answer no link or bot or I report you I report you if you report me
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  13. 2094.4
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  16. 37.7
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  18. 2065.24
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4 0
3 years ago
Sec s = 1.6948
Anastaziya [24]
That'd be true only if the value of "s" is the exact same one for both
namely  if sec(s) = cos(s)
then solving for "s"
thus

\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)}
\\\\\\
thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s)
\\\\\\
\pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides}
\\\\\\
cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s
5 0
3 years ago
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