To find the area of a trapezoid, dylan uses the formula a = 1/2 (b 1 + b 2)h the bases have lengths of 3.6 cm and 12 cm. the height of the trapezoid is cm.

the area of the trapezoid is irrational because
the values of the variables are all irrational numbers.
the entire answer is being multiplied by a fraction.
the height is irrational, and it is multiplied by the other rational dimensions.
the bases have an irrational sum that will be multiplied by the rational height.

the height is irrational, and it is multiplied by the other rational dimensions.

Step-by-step explanation:

Given that a trapezoid has bases as 3.6 cm and 12 1/3 cm.

But height is irrational as square root of 5 cm.

The area of the trapezoid is irrational because the area is found by adding the two bases and multiplying by half of height.

Sum of bases would be rational being sum of two rational numbers

But since height is irrational the area would be the product of a rational with irrational

Hence area would be irrational.

So correct answer is

Option c)the height is irrational, and it is multiplied by the other rational dimensions.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

2x^2+5x+17

your welcome. : )