Me with these all work is you: )❤️

3) In Δ BDC

|DC|/|BC| = cos C

cos C= 16/17.89

C= cos⁻¹( 16/17.89)=26.57⁰

In triangle ABC:

x=180-(90+26.57)=63.43

x=63.43

2)

AB/BD= tan(70⁰), AB=BD*tan(70⁰)

AB/BC=tan(40⁰), AB=BC*tan(40⁰)

BD*tan(70⁰)=BC*tan(40⁰)

BD=BC-CD=BC - 15

(BC -15)*tan(70⁰)=BC*tan(40⁰)

BC*tan(70⁰) -15*tan(70⁰)= BC*tan(40⁰)

BC*tan(70⁰) - BC*tan(40⁰) = 15*tan(70⁰)

BC(tan(70⁰)-tan(40⁰))= 15*tan(70⁰)

BC = 15*tan(70⁰)/(tan(70⁰)-tan(40⁰)) = 21.60

BC=21.60

1) In a quadrilateral sum of angles =360⁰.

PQS=SQR=50⁰, because SQ bisects PQR.

Using Law of sine in ΔSRQ

SR/sin ∠SQR = SQ/sinR, SQ = SR * sinR/sin SQR = 3*sin30/sin50 =3.26

SQ=3.26 cm

ΔPQS:

cos PQS= PQ/SQ

PQ=SQ*cosPQS =3.26*cos 50⁰=2.09=2.1

PQ=2.1 cm

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

4400

step-by-step explanation: