A) use synthetic division to show that 2 is a solution of 2h3 +14h2 − 72 = 0 .b) using part a, find the dimensions of a box that has a volume of 72 cubicinches where the width is twice the height and the length is 7 inches more than the height.4. find k so that 4x-3 is a factor of 20x3+23x2-10x+k

Length=9 inches,width=4 inches, height =2 inches.

k=

Step-by-step explanation:

a) If by synethetic division method the remainder equals i.e in the last row and last column if figure yield is 0 then 2 is one of the factor i.e one of the solution of given equation.

Remainder is 0.

Hence 2 is the solution of given equation.

b) Given a box having volume 72 cubic inches

   Let Height = x inches

   ∵ length is 7 inches more than the height

  ⇒     Length = x+7 inches

  & also width is twice the height

 ⇒      Width =

   Given   Volume = 72 cubic inches

    = 72

    - 72 = 0

From part a, 2 is the solution of above equation

⇒       Length = x+7 = 2+7 = 9 inches

          Width = = 4 inches

         Height = x = 2 inches

Given is a factor of

Hence,

                      k=

lol

step-by-step explanation:

∠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.