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Ad libitum [116K]

3 years ago

7

A(n.4) and B(6,8) and is parallel to y = 2x-5. What is the value of n?

1 answer:

agasfer [191]3 years ago

4 0

Answer:

i believe it is 4. the formula is 2x -4

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Simplify the expression

cestrela7 [59]

It would come out as 1

7 0

2 years ago

The cost (in dollars) of a custom made sweatshirt is represented by 3.5n +29.99, where n is the number of different colors in th

juin [17]

Answer:

$52.5n+449.85$

Step-by-step explanation:

The cost is given for <em>1 sweatshirt</em>. To find the cost of 15 sweatshirts, we need to multiply the expression by 15. Let's do this:

$15(3.5n+29.99)\\=15(3.5n+29.99)\\=15(3.5n)+15(29.99)\\=52.5n+449.85$

This is the expression that represents the cost of 15 sweatshirts.

8 0

3 years ago

Pls help me on this question

astraxan [27]

Answer:

h < 2

Step-by-step explanation:

Step 1: Distribute

10h + 40 < 60

Step 2: Subtract 40 on both sides

10h < 20

Step 3: Divide both sides by 10

h < 2

4 0

2 years ago

If tan θ =(√3)/3, 0◦< θ < 360◦,

svetlana [45]

Arctan (√3 /3) = 30°. = π/6 rad

That is the value searched, in degrees and radians.

You can verifiy that tan(30°) = sin(30°) / cos(30°) = [1/2] / [√3/2] = 1/√3 = √3 / 3

3 0

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