Find an ordered triple that represents 6y + 4z if y=(2,6,8) and z= (7,-2,6)

<40, 28, 72>

Step-by-step explanation:

y and z are the following vectors:

y = <2, 6, 8>

z = <7,  -2, 6>

To find 6y + 4z, first compute the scalar multiplications:

6y = <6*2, 6*6, 6*8> = <12, 36, 48>

4z = <4*7,  4*-2, 4*6> = <28,  -8, 24>

Then, add the obtained vectors, as follows:

6y + 4z = <12 + 28, 36 - 8, 48 + 24> =  <40, 28, 72>

the answer is 62

∠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.