The length of a rectangle is 1 more meter than three times its width. The rectangle has an area of 310 m².

Let w represent the width of the rectangle. What quadratic equation, in standard form, represents this situation?

Step-by-step explanation:

Let the width of rectangle be x m.

Given that length is 1 more than 3 times of times of width

Then width of rectangle in terms of x = 3x+1 (3 time width is 3x then added 1)

Area of rectangle is given by length * width

thus area of the given rectangle in terms of x = x*(3x+1) = 3x^2 +x

also area is given as 310 meter square.

Thus it can be said that

3x^2 +x = 310  subtracting 310 from both sides

=> 3x^2 +x -310 = 310 - 310

=> 3x^2 +x -310 = 0

Thus, quadratic equation, which  in standard form, represents this situation is given below

∠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.