PRS is isosceles with RP RS RQ is drawn such that it bisects PRS What additional fact can be used to prove PRQ SRQ by Sas in order to state that P S because they are congruent parts of congruent triangles? RQ /PS RQ /RQ PQR /SQR PQ /SQ

Step-by-step explanation:

An isosceles triangle has two equal sides and the two opposite angles to the sides to be equal.

Given that; RP = RS, RQ and PS are common,

RP = SQ (opposite sides of parallelogram RPQS)

PQ = RS (opposite sides of parallelogram RPQS)

ΔRPS = ΔQPS (congruence property)

Thus comparing triangles PQR and SQR,

= (similarity property)

<PRS = <PRQ + <SRQ (bisection of included <PRS)

<PQS = <PQR + <SQR (similatity property to <PRS)

 = (congruence property)

But, SRQ = PQR

So that;

Therefore by Side-Angle-Side (SAS), the required additional fact is:  

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

x= cost per bracelet

number of bracelets * x + shipping charge = total cost

so

9x+9=72