Which is the comparison between 0.731 and 0.73

Mathematics

2 answers:

Licemer1 [7]3 years ago

7 0

0.731 is greater than 0.73

NARA [144]3 years ago

4 0

<h3>0.731 is greater than 0.73 because number one is at the end which makes it greater. And look at it this way/ example: 7.401 or 7.041 which is 7.401 is greater than 7.041 because 7.041 stops at a Zero so u have to choose between those numbers and it would be 7.401 It is greater and it has a zero kind of like at the end~ ! hope dis helped!!! :) 0.73 is greater!</h3>

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I need help with it pls

Basile [38]

Answer:

Step-by-step explanation:

Step 1: What are the couch's original coordinates?

A: (-4, 2)
N: (-4, 3)
G: (-1, 3)
L: (-1, 4)
E: (-5, 4)
S: (-5, 2)

Step 2: Rotate the couch 90° counterclockwise.

Rule: (x, y) → (-y, x)

A': (-4, 2) → (-2, -4)
N': (-4, 3) → (-3, -4)
G': (-1, 3) → (-3, -1)
L': (-1, 4) → (-4, -1)
E': (-5, 4) → (-4, -5)
S': (-5, 2) → (-2, -5)

Step 3: Now reflect the "new" couch over the y-axis.

Rule: (x, y) → (-x, y)

A'': (-2, -4) → (2, -4)
N'': (-3, -4) → (3, -4)
G'': (-3, -1) → (3, -1)
L'': (-4, -1) → (4, -1)
E'': (-4, -5) → (4, -5)
S'': (-2, -5) → (2, -5)

Step 4: Finally translate the new new" couch right 1 unit and up 5 units.

Rule: (x + 1, y + 5)

A''': (2, -4) → (2 + 1, -4 + 5) → (3, 1)
N''': (3, -4) → (3 + 1, -4 + 5) → (4, 1)
G''': (3, -1) → (3 + 1, -1 + 5) → (4, 4)
L''': (4, -1) → (4 + 1, -1 + 5) → (5, 4)
E''': (4, -5) → (4 + 1, -5 + 5) → (5, 0)
S''': (2, -5) → (2 + 1, -5 + 5) → (3, 0)

Step 5: Use the distance formula to show that the length of EG is the same as the length of E'''G'''.

EG: (-5, 4)(-1, 3)
E'''G''': (5, 0)(4, 4)

Distance between E(-5, 4) and G(-1, 3)

x₁ = -5
x₂ = -1
y₁ = 4
y₂ = 3

$\large \textsf {d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$}\\\\\large \textsf {d = $\sqrt{(-1-(-5))^2+(3-4)^2}$}\\\\\large \textsf {d = $\sqrt{(-1+5)^2+(3-4)^2}$}\\\\\large \textsf {d = $\sqrt{4^2+(-1)^2}$}\\\\\ \large \textsf {d = $\sqrt{16+1}$}\\\\\large \textsf {d = $\sqrt{17}$}\\\\\large \textsf {d = ${4.12}$}$

Distance between E'''(5, 0) and G'''(4, 4)

x₁ = 5
x₂ = 4
y₁ = 0
y₂ = 4

$\large \textsf {d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$}\\\\\large \textsf {d = $\sqrt{(4-5)^2+(4-0)^2}$}\\\\\large \textsf {d = $\sqrt{(-1)^2+4^2}$}\\\\\large \textsf {d = $\sqrt{1+16}$}\\\\\ \large \textsf {d = $\sqrt{17}$}\\\\ \large \textsf {d = ${4.12}$}$