Claire and Richard are both artists who use square canvases. Claire uses the polynomial 50%? + 250 to decide how much to charge for her paintings

and Richard uses the polynomial 40x² + 350 to decide how much to charge for

his paintings. In each polynomial, x is the height of the painting in feet.

a. How much does Claire charge for a 20-foot-tall painting?

b. How much does Richard charge for a 15-foot-tall painting?

c. To the nearest tenth, for what height will both Claire and Richard charge

the same amount for a painting? Explain how to find the answer.

d. When both Claire and Richard charge the same amount for a painting,

how much does each charge?

A.  

B.  

C.    Height = 3.2 feet

D. Charges = $760

Step-by-step explanation:

Given

Solving (a): Claire's 20ft charges

In this case, x = 20

Substitute 20 for x in

Solving (b): Richard's 15ft charges

In this case, x = 15

Substitute 20 for x in

Solving (c): Height which they both charge the same;

This implies that

Solving for x [Collect Like Terms\

Divide both sides by 10

Take square roots of both sides

 (Approximated)

Hence; Height = 3.2 feet

Solving (d): How much they charge when the charge the same amount.

Substitute 3.2 for x in any of the given equation

The reason for the difference is due to approximation

Hence, they both charge approximately 760

e) none of the above because the measure of angle cba is 101°

∠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.