Answer:
Step-by-step explanation:
Answer
S=3
Step-by-step explanation:
add 24 to both sides, then you subtract -42+24 take sign of the larges which is negative and youll get -6s=-18 divide -6 to both sides and you will get S=3
Answer:
the original volume is lesser than the new volume by 242.47 in³
Step-by-step explanation:
Volume of a cone is;
V = ⅓πr²h
Where;
r is radius
h is height
For the original chamber;
r = 5.7/2 = 2.85 inches
h = 12 inches
Volume of this original chamber is;
V_orig = ⅓ × π × 2.85² × 12
V_orig = 102.02 in³
In the new design, the chamber is scaled by a factor of 1.5.
Thus;
r_new = 2.85 × 1.5 = 4.275 inches
h_new = 12 × 1.5 = 18 inches
V_new = ⅓ × π × 4.275² × 18
V_new = 344.49 inch³
V_new - V_orig = 344.49 - 102.02 =
Thus, the original volume is lesser than the new volume by 242.47 in³
Answer:
Part 1) m∠1 =(1/2)[arc SP+arc QR]
Part 2) 
Part 3) PQ=PR
Part 4) m∠QPT=(1/2)[arc QT-arc QS]
Step-by-step explanation:
Part 1)
we know that
The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.
we have
m∠1 -----> is the inner angle
The arcs that comprise it and its opposite are arc SP and arc QR
so
m∠1 =(1/2)[arc SP+arc QR]
Part 2)
we know that
The <u>Intersecting Secant-Tangent Theorem,</u> states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
so
In this problem we have that

Part 3)
we know that
The <u>Tangent-Tangent Theorem</u> states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments
so
In this problem
PQ=PR
Part 4)
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
In this problem
m∠QPT -----> is the outer angle
The arcs that it encompasses are arc QT and arc QS
therefore
m∠QPT=(1/2)[arc QT-arc QS]