You are given two pairs of triangles. for the first pair of triangles, each angle and side of one triangle is congruent to the corresponding angle and side of the other. you show that rigid motions can transform one triangle so that it matches up with the other. for the second pair of triangles, you show that rigid motions can transform one triangle so that each side or angle of one triangle matches exactly with a corresponding side or angle of the other triangle. what have you proved?

answer:

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

explanation:

service only with 20% tip  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

in the opening of the question the word 'service' is associated only with the pretax value.

the closing part of the question asks: how much did she pay for the  

service

−−−−−−  

so technically the answer is  

→

$

72.6

(

1

+

20

100

)

=

$

87.12

however, the intention of the question provider may have been that your answer is to be  

$

87.12

(

1

+

8

100

)

=

$

94.09

to 2 decimal places.

'

payment just for service  

→

$

87.12

payment for service including tax

→

$

94.09

tony b

mar 21, 2017

answer:

trying to explain it another way

explanation:

service charge  

.

→

$

72.60

20% tip  

20

100

×

$

72.60

=

$

14.52

−−−−−−  

←

add

$

87.12

tax  

8

100

×

$

87.12

=

$

6.97

−−−−−  

←

add

$

94.09

related questions

∠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.

where's the diagram?

step-by-step explanation:

b

step-by-step explanation: