All categories

3 years ago

How many rows are needed to seat all students. When there are 846 students and each row holds 18 people

Mathematics

1 answer:

lubasha [3.4K]3 years ago

6 0

Answer:

47 rows

Step-by-step explanation:

To find the amount of needed rows, we will use the math operation division. We will take the total and divide it by the amount in each row.

846/18=47

To seat 18 people in each row, you will need 47 rows.

You might be interested in

1.)Find the height of a rectangular prism if the surface area is 148 cm2, the width is 4 cm and the length is 5 cm.

Contact [7]

1) surface of a rectangular prism=2(length x width)+2(length x height)+2(width x height)

Therefore:
148 cm²=2(5 cm x 4 cm)+2(5 cm x h)+2(4 cm x h)=
148 cm²=40 cm²+10 cm h+8 cm h
18 cm h=148 cm²-40 cm²
18 cm h=108 cm²
h=108 cm² / 18 cm=6 cm.

answer: height=6 cm

2)
Volume of a rectangular prism= length x width x height
therefore:
34 cm³=(1.7 cm)(0.5 cm) h
0.85 cm² h=34 cm³
h=34 cm³/0.85 cm²
h=40 cm.

answer: height=40 cm

3)
volume of a cylinder: πr²h
therefore.
118.79 ft³=πr²(5 ft)
r=√(118.79 ft³/5π ft)≈2.75 ft

answer: radius=2.75 ft

4)

Surface area of the pyramid with square base=4(A side)+A base
A side=(1/2)(8ft)(12 ft)=48 ft²
A base=(8 ft)(8 ft)=64 ft²

surface area=4(48 ft²)+64 ft²=256 ft²

Answer: the surface area of this pyramid would be 256 ft².

5)
surface of a cone=πrs+πr²
therefore:
radius=diameter/2=6.2 ft/2=3.1 ft
63.3 ft²=π(3.1 ft) s+π(3.1 ft)²
3.1π ft s=33.109 ft²
s=33.109 ft² /3.1π ft
s≈3.4 ft

Answer: the slant height would be 3.4 ft.

6)
volume of a square pyramid=(area of base x heigth)/3
therefore:
area of base=(6 ft)(6 ft)=36 ft²
126.97 ft³=36 ft² h /3
h=126.97 ft³/12 ft²=10.58 ft

answer: the height would be 10.58 ft.

7)
volume of a cone =(base x height)/3
base of a cone=πr²
therefore:
199.23 cm³=πr²(9 cm)/3
r=√(199.23 cm³ / 3π cm)≈4.6 cm

answer: the radius would be 4.6 cm.

5 0

3 years ago

Answer all three questions

inn [45]

Answer:

3 . al of this

4 . keeping in heat

3 0

3 years ago

Triangle MRN is created when an equilateral triangle is folded in half.

Rus_ich [418]

The value of y based on the values given ib the equilateral triangle will be 4.

<h3>How to illustrate the triangle?</h3>

From the information given, MN is one of the sides of the equilateral triangle. Therefore, side MR will be:

= 1/2 × 8 = 4

In the above, it should be noted that MR was calculated since it's half of the value of MN which is 8.

Here, value of y based on the values given ib the equilateral triangle will be 4.

Learn more about triangles on:

brainly.com/question/17335144

#SPJ1

6 0

2 years ago

Do any of you get this question? Please help!

Alexxandr [17]

Honestly i just wanted to try this question since ive never seen it but i dont rlly know if i did it right at all

I equaled GH, HI, and GI together to get y

which i got y=-2

and what i got next i feel is off since

GH, HI, and GI all equaled -11

if you kind of know how to do the question maube you could correct me from there but otherwise dont take my word for it completely

4 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers