What’s the gcf of 18k and 15k cubed?

The GCF for the variable part is  

k

Step-by-step explanation:

Since  

18

k

,

15

k

3

contain both numbers and variables, there are two steps to find the GCF (HCF). Find GCF for the numeric part then find GCF for the variable part.

Steps to find the GCF for  

18

k

,

15

k

3

:

1. Find the GCF for the numerical part  

18

,

15

2. Find the GCF for the variable part  

k

1

,

k

3

3. Multiply the values together

Find the common factors for the numerical part:

18

,

15

The factors for  

18

are  

1

,

2

,

3

,

6

,

9

,

18

.

Tap for more steps...

1

,

2

,

3

,

6

,

9

,

18

The factors for  

15

are  

1

,

3

,

5

,

15

.

Tap for more steps...

1

,

3

,

5

,

15

List all the factors for  

18

,

15

to find the common factors.

18

:  

1

,

2

,

3

,

6

,

9

,

18

15

:  

1

,

3

,

5

,

15

The common factors for  

18

,

15

are  

1

,

3

.

1

,

3

The GCF for the numerical part is  

3

.

GCF

Numerical

=

3

Next, find the common factors for the variable part:

k

,

k

3

The factor for  

k

1

is  

k

itself.

k

The factors for  

k

3

are  

k

⋅

k

⋅

k

.

k

⋅

k

⋅

k

List all the factors for  

k

1

,

k

3

to find the common factors.

k

1

=

k

k

3

=

k

⋅

k

⋅

k

The common factor for the variables  

k

1

,

k

3

is  

k

.

k

The GCF for the variable part is  

k

.

GCF

Variable

=

k

Multiply the GCF of the numerical part  

3

and the GCF of the variable part  

k

.

3

k

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.