Find the area of the polygon XYZ that has its vertices at X(–3, 6), Y(–3, 1), and Z(5,1). Question 8 options: A) 26 square units B) 20 square units C) 40 square units D) 6.5 square units

Area ≈ 20 square units

Step-by-step explanation:

Using Distance Formula to Find the lengths

Distance Formula =

Length XY:

=>

=>

=> 5 units

Length YZ:

=>

=>

=> 8 units

Length ZX:

=>

=>

=> 9.4

Perimeter:

=> 5+8+9.4

=> 22.4

Semi-Perimeter:

=> 11.2

Using Heron's Formula to find the area:

Area =

Where s is semi perimeter and a,b and c are side lengths

=> Area =

=> Area =

=> Area =

=> Area = 19.99

=> Area ≈ 20 square units

answer: 2/4

step-by-step explanation:

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!