A circle with radius of \greenD{1\,\text{cm}}1cmstart color greenD, 1, space, c, m, end color greenD sits inside a \blueD{11\,\text{cm} \times 12\,\text{cm}}11cm×12cmstart color blueD, 11, space, c, m, times, 12, space, c, m, end color blueD rectangle.

What is the area of the shaded region?

Round your final answer to the nearest hundredth.

∠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.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

i think it might be a but you might want a second opinion