In ΔCDE, \text{m}\angle C = (5x+18)^{\circ}m∠C=(5x+18) ∘ , \text{m}\angle D = (3x+2)^{\circ}m∠D=(3x+2) ∘ , and \text{m}\angle E = (x+16)^{\circ}m∠E=(x+16) ∘ . Find \text{m}\angle D. M∠D.

Angle D = m∠D= 50°

Step-by-step explanation:

In ΔCDE we are given: 3 angles

m∠C=(5x+18) ∘

m∠D=(3x+2) ∘

m∠E=(x+16) ∘ .

The sum of angles in a triangle = 180°

Hence:

m∠C + m∠D + m∠E = 180°

(5x+18)° + (3x+2)° + (x+16)° = 180°

5x + 18 + 3x + 2 + x + 16 = 180°

5x + 3x + x + 18 + 2 + 16 = 180°

9x +36= 180°

Subtract 36 from both sides

9x + 36 - 36 = 180° - 36°

9x = 144°

x = 144°/9

x = 16

From the above question, we are asked to find:angle D (m∠D )

Hence:

m∠D=(3x+2)°

m∠D=( 3 × 16 + 2)°

m∠D=(48 + 2)°

m∠D= 50°

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!