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3 years ago

6

Write an equation of a line that is parallel to the given line.

1 answer:

Marina CMI [18]3 years ago

6 0

Answer:

5) y=5

6)x=7

7)a=18

Step-by-step explanation:

5) y= any number will work. don't add a variable like x

6)x= any number will work. don't add a variable

My work for 7 is in the picture. If you'd like me to explain more for 7 lmk :)

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Mr.Willams' physical education class lasts 7/8 hours. How many minutes are spent warming up and cooling down?

Ray Of Light [21]

7/8 of an hour is 52.5 minutes. 3/10 of 52.5 minutes is 15.75 minutes. They spent 15.75 minutes warming up and cooling down.

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3 years ago

Regular pentagon RSTUV has sides that are half the length of regular pentagon JKLMN.

Andrew [12]

Answer:

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Step-by-step explanation:

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2 years ago

Divide the whole into pieces that are each ⅙ of the whole. Then, solve the equation.

yKpoI14uk [10]

Answer: 6

Explanation

If we divide our figure into six, we get the following figure:

As we can see, we are left with 6 equal parts. This can be confirmed by solving the equation:

$1\div\frac{1}{6}=\frac{6}{1}=6$

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1 year ago

Find the value of x and the value of y

WARRIOR [948]

I’m pretty sure it’s b.

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3 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