All categories

3 years ago

In how many different ways can seven children line up in one row to have their picture taken?

Mathematics

2 answers:

Citrus2011 [14]3 years ago

8 0

Answer:

5,040!

Step-by-step explanation:

I know you said it below, don't think im copying i'm just putting it here so other people can see it more easily!

Anyways have a lovely Day/night and goodluck on your studies!! <3

____ [38]3 years ago

6 0

I think it’s 49 cause it’s definitely more than 7 and the other two options seem far too high and unreasonable so it’s most likely 49 :)

You might be interested in

A gardener transplanted a plant from an indoor pot to an outdoor garden. After 30 days, the plant's height was 18 centimeters. A

Aliun [14]

Let x is the days and the y is the height of the plant.

Let the change between x and y is linear.

So, the relation between x and y will take the form ⇒ y = ax + b

where a and b are constants.

At x = 30 ⇒ y = 18 ⇒⇒⇒ ∴ 18 = 30 a + b → (1)

At x = 90 ⇒ y = 29 ⇒⇒⇒ ∴ 29 = 90 a + b → (2)

solve (1) and (2) to find a and b

subtract (1) from (2)

∴ 11 = 60 a ⇒⇒⇒ ∴ a = 11/60

substitute at (1) ⇒⇒⇒ ∴ b = 25/2 = 12.5

So, the equation which represents the plants growth is

∴ y = (11/60) x + 12.5

4 0

3 years ago

PLZ HELP I NEED THIS IN 5 MINS (23+3x)+(8x–41)=15

swat32

Hello from MrBillDoesMath!

Answer:

x = 3

Discussion:

( 23 + 3x) + (8x-41) = 15 =>

23 - 41 + 3x + 8x = 15 =>

-18 + 11x = 15 =>

11x = 15 + 18 = 33 =>

x = 33/11 = 3

Thank you,

MrB

7 0

3 years ago

Read 2 more answers

△ABC ~ △DEF Find the length of BC . 72 54 40.5 108

tatiyna

Answer:

The answer is 72

7 0

4 years ago

Read 2 more answers

9. Plot the points A (-2,3), B(2,-3), C(2,3) ,D(-2,3) join the points and find the distance between each point.

pav-90 [236]

Answer:

AB=7.21 unit

BC=6 unit

CD=7.21 unit

AD= 6 unit

AC=4 unit

BD=4 unit

Step-by-step explanation:

Coordinates of A =(-2,3)

Coordinates of B = (2,-3)

Coordinates of C = (2,3)

Coordinates of D =(-2,-3)

Distance formula : $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$A=(x_1,y_1)=(-2,3)\\B=(x_2,y_2)=(2,-3)\\AB = \sqrt{(2+2)^2+(-3-3)^2}$

AB=7.21 unit

$B=(x_1,y_1)=(2,-3)\\C=(x_2,y_2)=(2,3)\\BC=\sqrt{(2-2)^2+(3+3)^2}$

BC=6

$C=(x_1,y_1)=(2,3)\\D=(x_2,y_2)=(-2,-3)\\CD=\sqrt{(-2-2)^2+(-3-3)^2}$

CD=7.21

$A=(x_1,y_1)=(-2,3)\\D=(x_2,y_2)=(-2,-3)\\AD=\sqrt{(-2+2)^2+(-3-3)^2}$

AD=6

$A=(x_1,y_1)=(-2,3)\\C=(x_2,y_2)=(2,3)$

$AC=\sqrt{(2+2)^2+(3-3)^2}$

AC=4

$B=(x_1,y_1)=(2,-3)\\D=(x_2,y_2)=(-2,-3)$

BD= $\sqrt{(-2-2)^2+(-3+3)^2}$

BD=4

4 0

3 years ago

You have a 1200-foot roll of fencing and a large field. you want to make two paddocks by splitting a rectangular enclosure in ha

romanna [79]

Let width = w
Let length = l
Let area = A
3w+2l=1200
2l=1200-3w
l=1200-3/2
A=w*l
A=w*(1200-3w)/2
A=600w-(3/2)*w^2
If I set A=0 to find the roots, the maximum will be at wmax=-b/2a which is exactly 1/2 way between the roots-(3/2)*w^2+600w=0
-b=-600
2a=-3
-b/2a=-600/-3
-600/-3=200
w=200
And, since 3w+2l=1200
3*200+2l=1200
2l = 600
l = 300
The dimensions of the largest enclosure willbe when width = 200 ft and length = 300 ft
check answer:
3w+2l=1200
3*200+2*300=1200
600+600=1200
1200=1200
and A=w*l
A=200*300
A=60000 ft2
To see if this is max area change w and l slightly but still make 3w+2l=1200 true, like
w=200.1
l=299.85
A=299.85*200.1
A=59999.985

8 0

4 years ago