Atriangle has sides qp and rq of triangle pqr are produced to points s and t respectively if< spr=135° and < pqt=110°, find < prq

∠ PRQ = 65°

Step-by-step explanation:

See the diagram attached.

The ∠ PQR and ∠ PQT are supplementary.

So, ∠ PQR = 180° - 110° = 70° {Since given that ∠ PQT = 110°}

Again, ∠ SPR and ∠ RPQ are supplementary.

So, ∠ RPQ = 180° - 135° = 45° {Since, given that ∠ SPR = 135°}

Now, from Δ PQR, ∠ PQR + ∠ RPQ + ∠ PRQ = 180°

⇒ 70° + 45° + ∠ PRQ = 180°

⇒ ∠ PRQ = 180° - 70° - 45° = 65° (Answer)

answer: the answer is 5/22

step-by-step explanation: if you subtract 17 from 22 the answer is 5 therefore the remainder of students are female

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!