Answer:
Three cubes
Step-by-step explanation:
The cubes have to be indistinguishable and all orientations of one cube are also have to be indistinguishable.
All ways of connecting two cubes result in the same shape. So answer is larger than two.
After connecting two cubes, there are ten faces where the third cube can be attached, and two faces which are connected, accounting for all 12 faces of two cubes.
Of the 10 exposed faces, exactly two are on opposite ends, both leading to the same straight line figure. The other 8 faces all lead to an L shape, and all L shapes can be rotated to be identical.
Hence, three cubes can only make a straight shape or an angled shape.
Four cubes can make a straight shape, a L shape, a Γ shape (but flipping it over through 3 dimensions makes L and Γ identical), a T shape, and a square shape. That is either four or five different objects depending on if they can be lifted from the table. Anyway, it is more than two.
Multiply both sides of the second equation by 100 to get rid of the decimals:
0.05<em>n</em> + 0.10<em>d</em> = 1.50
==> 5<em>n</em> + 10<em>d</em> = 150
Multiply both sides of the first equation by -5:
<em>n</em> + <em>d</em> = 21
==> -5<em>n</em> - 5<em>d</em> = -105
Add the two equations together:
(5<em>n</em> + 10<em>d</em>) + (-5<em>n</em> - 5<em>d</em>) = 150 + (-105)
Notice that the terms containing <em>n</em> get eliminated and we can solve for <em>d</em> :
(5<em>n</em> - 5<em>n</em>) + (10<em>d</em> - 5<em>d</em>) = 150 - 105
5<em>d</em> = 45
<em>d</em> = 45/5 = 9
Plug this into either original equation to solve for <em>n</em>. Doing this with the first equation is easiest:
<em>n</em> + 9 = 21
<em>n</em> = 21 - 9 = 12
So Donna used 12 nickels and 9 dimes.
Answer:
This will be letter j). 1,767.1 in 3
Step-by-step explanation:this is the closest estimate to the cubic inch plz brainliest
1/6 = 0.16 with a line over the 6 because it is repeating
Answer:
C
Step-by-step explanation:
5+2y+3z=5+3z+2yA
Intercambie los lados para que todos los términos de las variables estén en el lado izquierdo.
5+3z+2yA=5+2y+3z
Resta 5 en los dos lados.
3z+2yA=5+2y+3z−5
Resta 5 de 5 para obtener 0.
3z+2yA=2y+3z
Resta 3z en los dos lados.
2yA=2y+3z−3z
Combina 3z y −3z para obtener 0.
2yA=2y
Anula 2 en ambos lados.
yA=y
Divide los dos lados por y.
y
yA
=
y
y
Al dividir por y, se deshace la multiplicación por y.
A=
y
y
Divide y por y.
A=1
