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Troyanec [42]
3 years ago
11

Find the approximate side length of a square game board with an area of 186in2

Mathematics
2 answers:
Arturiano [62]3 years ago
8 0
13.6in is the answer
abruzzese [7]3 years ago
7 0
I’m pretty sure it’s 13.6
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Sedaia [141]

The inverse is where the x and y values are flipped, so the left side would have 6 7 8 9 and the right side would have 9 10 11 12 for the inverse.

Therefore your answer is A.

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3 years ago
The model builder has 4 pieces of balsa wood that are 4 cm, 5 cm, 6 cm, and 7 cm length. How many different combinations of 3 pi
s2008m [1.1K]
About 22 i think hope it helps in any way

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3 years ago
Which is the MOST a Which cell structure functions like a power plant for a eukaryotic cell? A. nucleus B. cytoplasm C. mitochon
Ivahew [28]

Answer:

c

Step-by-step explanation:

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3 years ago
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Find the value of the expression m − 4.6 for m = 7.
stich3 [128]

Answer:

2.4

Step-by-step explanation:

If you replace m with 7, the expression is now 7-4.6.

7-4.6=2.4

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-hope it helps

3 0
2 years ago
A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
Alex_Xolod [135]

Answer:

The number of seat when n is odd S_n=\frac{n^2+2n+1}{4}

The number of seat when n is even S_n=\frac{n^2+2n}{4}

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

S_n=\frac{n}{2}[2a+(n-1)d]

    =\frac{n}{2}[a+l]

a = first term of the series.

d= common difference.

n= number of term

l= last term

n^{th} term of a A.P series is

T_n=a+(n-1)d

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=1 and d=2

n=1+(t-1)2

⇒(t-1)2=n-1

\Rightarrow t-1=\frac{n-1}{2}

\Rightarrow t = \frac{n-1}{2}+1

\Rightarrow t = \frac{n-1+2}{2}

\Rightarrow t = \frac{n+1}{2}

Last term l= n,, the number of term =\frac{ n+1}2, First term = 1

Total number of seat

S_n=\frac{\frac{n+1}{2}}{2}[1+n}]

    =\frac{{n+1}}{4}[1+n}]

     =\frac{(1+n)^2}{4}

    =\frac{n^2+2n+1}{4}

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let t^{th} of the series is n.

T_n=a+(n-1)d

Here T_n=n, n=t, a=2 and d=2

n=2+(t-1)2

⇒(t-1)2=n-2

\Rightarrow t-1=\frac{n-2}{2}

\Rightarrow t = \frac{n-2}{2}+1

\Rightarrow t = \frac{n-2+2}{2}

\Rightarrow t = \frac{n}{2}

Last term l= n, the number of term =\frac n2, First term = 2

Total number of seat

S_n=\frac{\frac{n}{2}}{2}[2+n}]

    =\frac{{n}}{4}[2+n}]

     =\frac{n(2+n)}{4}

    =\frac{n^2+2n}{4}  

4 0
2 years ago
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