At the end of a 2000 - 2001 football season 31 super bowl games had been played with the current two football league's the american football conference afc and the national football conference nfc the nfc won five more games in the afc determine the total number of wins by each conference

American Football Conference   (AFC) won  =  x  = 13

National Football Conference   (NFC)   won  =  y  = 18

Step-by-step explanation:

Total number of Super Bowl games   31   at the end of 2000-2001 season

American Football Conference   (AFC)  , and National Football Conference

Let   x  be number of Super Bowl won by AFC

Let   y  be number  of Super Bowl won by NFC

According to problem statement

y  = x + 5    (1)

Then we got a  two equation system with

x  +  y  =  31   (2)

We solve such system, by subtitution  (value of  y  from equation (1) in equation (2)

x  + ( x + 5 )  =  31

2x  +  5  =  31

2x   = 31 -5

2x  =  26

x  = 13

and   y =  18

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.