Answer:
125
Step-by-step explanation:
that would be , by proportion. 5 * 800 / 32
= 125 students
Answer:
175°
Step-by-step explanation:
Bearing angles are usually measured clockwise from North. Reverse bearing angles differ from forward bearing angles by 180°. These relations and the usual angle sum relation for a triangle can be used to solve this problem.
Angle PQR will be the difference in the bearings from Q to P and Q to R:
∠PQR = 124° -46° = 78°
Triangle PQR is isosceles, so the base angle at P will be ...
∠QPR = (180° -78°)/2 = 51°
__
The bearing from P to R will be 51° less than the bearing from P to Q. The bearing from P to Q is 180° more than the bearing from Q to P.
PR bearing = PQ bearing - ∠QPR
= PQ bearing - 51°
= (46° +180°) -51° = 175°
The bearing of R from P is 175°.
Answer:
-7
Step-by-step explanation:
y=3x-7
x=0 ⇒ y= 3*0-7= -7
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
Answer:
8.
Step-by-step explanation:
5, 10, 12, 4, 6, 11, 13, 5
Arrange in ascending order:
4, 5, 5, 6, 10, 11, 12, 13.
The median is the mean of the 2 middle numbers:
= (6 + 10) / 2
= 8.
