Find the angle measure indicated. assume that lines which appear to be tangent are tangent.

( answer both and i'll give you brainliest! )

13)  135°   14)  143°

Solution:

13) The lengths of tangents drawn from an external point are equal.

So, PA = PB

Tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.

OB ⊥ PB and OA ⊥ PA

So, ∠PBO = 90°, ∠PAO = 90°

Sum of the angles in a quadrilateral is 360°.

In quadrilateral PAOB,

⇒ ∠APB + ∠PAO + ∠PBO + ∠AOB = 360°

⇒ 45° + 90° + 90° + ∠AOB = 360°

⇒ ∠AOB = 360° – 225°

⇒ ∠AOB = 135°

Hence the missing angle is 135°

14)  Angle in a semi-circle is 90° .

∠ACB = 90°

Sum of the angles in triangle is 180° .

In ∆ABC, ∠ACB + ∠ABC + ∠CAB = 180°

⇒ 90° + 53° + ∠CAB = 180°

⇒ ∠CAB = 180° – 143°

⇒ ∠CAB = 37°

AD and DE are radius.

So, ∠DEA = 37°

Sum of the adjacent angles in a straight line is 180°.

⇒ ∠DEA + ∠DEB = 180°

⇒ 37° + ∠DEB = 180°

⇒ ∠DEB = 143°

Hence the missing angle is 143°

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!

7,24

step-by-step explanation: