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alexandr1967 [171]

3 years ago

7

What is the volume of this rectangular prism?

2 answers:

Marina CMI [18]3 years ago

5 0

206.25 or 205 1/2 all that was needed to be done was multiply all three sides by each other

valkas [14]3 years ago

3 0

Answer: 206 1/4 inches (or 206.25in)

to find the volume you do
length x width x height
so
7 1/2 x 2 1/2 get your sum
then multiply that by your height.

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If 6 people shared 5 blocks of clay evenly, how much would everyone get?

Lena [83]

Divide the number of blocks by number of people In this case 5/6 = 0.83 blocks or each person will get 0.83 or 5/6 of a block

5 0

3 years ago

A baker decorates 42cupcakes in 30 minutes. She decorates cupcakes at a constant rate.

eimsori [14]

Answer: 1.4 min per min

Step-by-step explanation: sorry if wrong

8 0

2 years ago

The zeros of function dare -3 and 8. Which expression could be a factor of function d?

erastovalidia [21]

Answer:

(x + 3)

Step-by-step explanation:

Using the zero product property

x-a = 0 x-b = 0 where a and b are the zeros

(x-a)(x-b) =0

(x- -3)(x -8) =0

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3 0

3 years ago

Will give brainliest if correct

GenaCL600 [577]

Answer: put the parallelograms in the box that says 2 pairs of parallel sides next put the squares in the box that says 4 right angles the put rectangles in the box that says 4 congruent sides and last put rhombuses in the middle box that says 4 right angles and 4 congruent sides

Step-by-step explanation:

5 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