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3 years ago

Drag each number to the correct location on the table.

Mathematics

2 answers:

ASHA 777 [7]3 years ago

7 0

Answer:

$\text{Rational number} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ Irrational number}$

$3\sqrt{25} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{141}\\\\\sqrt{256} \\\\1. \overline{ 1625} \\\\\frac{-11}{151}$

Step-by-step explanation:

Since we believe it

The number sequence is that those which can be described as $\frac{p}{q}$ and that is not ending and recurring. Figures that are unreasonable are also the ones, that are not $\frac{p}{q}$ and therefore are non-ending, non-recurring.

The number sequence is $3\sqrt{25} = 3 \times 5 = 15$

Rational amount since it is classified as $\frac{p}{q}$ . $\frac{-11}{151}$ is really a rational number.

$1.\overline{1.625}$ is a real function, since it does not end but repeats it.

The rational number $\sqrt {256} = 16$

Unreasonable number $\sqrt{141}$ .

iogann1982 [59]3 years ago

4 0

Answer:

Since we believe it

The number sequence is that those which can be described as and that is not ending and recurring. Figures that are unreasonable are also the ones, that are not and therefore are non-ending, non-recurring.

The number sequence is

Rational amount since it is classified as . is really a rational number.

is a real function, since it does not end but repeats it.

The rational number

Unreasonable number .

Step-by-step explanation:

i used my brain

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So the answer is x=2

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3 years ago

A line segment has a midpoint of (5, 1/2 )<br> If one endpoint is (5,4), what is the other endpoint?

garik1379 [7]

Answer:

Step-by-step explanation:

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3 years ago

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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

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

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

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

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kondor19780726 [428]

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