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Eddi Din [679]
3 years ago
15

Which of the following is an equivalent form of the compound inequality

Mathematics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

pls mark brainy

Step-by-step explanation:

option (d) is correct.

An equivalent form of the  given compound inequality −44 > −2x − 8 ≥ −8 is  −44 > −2x − 8 and −2x − 8 ≥ −8

Step-by-step explanation:

Given a compound inequality −44 > −2x − 8 ≥ −8

We have to write  an equivalent form of compound inequality.

Compound inequality  consists of two inequalities joined together and the solution is the intersection of each inequality.

Compound inequality has two sides the left hand side and right hand side we can solve them by taking each inequality one at a time.

For given compound inequality, −44 > −2x − 8 ≥ −8

we have

Left side of inequality as  −44 > −2x − 8

and right side of inequality as  −2x − 8 ≥ −8

Thus, option (d) is correct.

Thus, An equivalent form of the  given compound inequality −44 > −2x − 8 ≥ −8 is  −44 > −2x − 8 and −2x − 8 ≥ −8

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the angle between a tangent and the diameter = 90°

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A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece
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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}  

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