Answer:
about 13.50$
Step-by-step explanation:
Plz name me the brainlest
Answer:
A.
Step-by-step explanation:
Since the lines are parallel and you know the slope of

is -2, you know the slope of line

will also be -2. You can see on the coordinate grid that line

crosses the y-axis at -1. When you piece all of these facts together into slope-intercept form you get
= -2x - 1.
Answer:
The answer is 675.75 square Inches
Step-by-step explanation:
<h3><u>Given</u>;</h3>
<h3>
<u>To </u><u>Find</u>;</h3>
<h3>
<u>Formula</u>;</h3>
Now,
Radius (r) = diameter ÷ 2
r = 14/2 in.
r = 7 in.
Now,
A = 4πr²
A = 4 × 22/7 × (7)² in.
A = 4 × 22 × 7 in.
A = 615.75 square in.
Thus, The area of the beach ball with a diameter of 14 in. is 615.75 square in.
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]