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Free_Kalibri [48]

2 years ago

9

Can anyone solve this?

1 answer:

podryga [215]2 years ago

4 0

Answer:

The answer is 1

Step-by-step explanation:

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Point A (4,-3) is reflected over the y- axis. What are<br> the coordinates of A'?

yan [13]

Answer:

A'(-4,-3)

Step-by-step explanation:

use formula of reflection over y axis

(x,y)=(-x,y)

A(4,-3)=A'(-4,-3)

5 0

2 years ago

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What is a rational number between 6.2 and 6.3

Nikitich [7]

.1 is the rational number

3 0

2 years ago

I don’t know how to do this question pls help

sineoko [7]

Answer:

Step-by-step explanation:

Triangle P is mapped onto Q so P is the initial triangle that will transformed.

We can rotate counterclockwise 90° but we cannot do it about the origin (0,0) because the red point (5, 1) will end up at ( -1, 5) .

We see that the point (5, 1) ends up at ( -2, 4) so the center of rotation is lower than the origin.

The transformation is rotation of 90° about the point (0,-1)

5 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

3 years ago

Which set of order pairs

REY [17]

The correct answer is A because the 4 in the X value repeats two time and on the other ones doesn't

8 0

2 years ago

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