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3 years ago

Solve the equation show you work and check. Your answer 4y+5=31

2 answers:

8 0

To find y, which is what I believe you mean by solving the equation, you have to first subtract 5 from each side. You'll have

4y=26

Then you have to divide each side by 4. Then you will get

y= 6.5

taurus [48]3 years ago

3 0

31-5=26
26/4=6.5
Y=6.5

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Which coordinate pair is the best estimate of the point of intersection in this graph?

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C o.75,0.5 yes it is

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A store owner puts in an order for product x and product y for a total of 4,000 units. it costs $0.10 per item to ship of produc

galben [10]

The system of equations can be used to determine how much of product x and product y the store owner bought is x + y = 4,000

0.10x + 0.04y = 352

<h3>Simultaneous equation</h3>

product x
Product y
Total units of x and y = 4,000 units
Cost of shipping each product x = $0.10
Cost of shipping each product y = $0.04
Total cost of shipping = $352

The equation:

x + y = 4,000

0.10x + 0.04y = 352

Therefore, the system of equations can be used to determine how much of product x and product y the store owner bought is x + y = 4,000

0.10x + 0.04y = 352

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2 years ago

Simplify the expression 3 (5x)

Dovator [93]

Answer:

15x

Step-by-step explanation:

i just know (100% positive)

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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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