Strange question, as normally we would not calculate the "area of the tire." A tire has a cross-sectional area, true, but we don't know the outside radius of the tire when it's mounted on the wheel.
We could certainly calculate the area of a circle with radius 8 inches; it's
A = πr^2, or (here) A = π (8 in)^2 = 64π in^2.
The circumference of the wheel (of radius 8 in) is C = 2π*r, or 16π in.
The numerical difference between 64π and 16π is 48π; this makes no sense because we cannot compare area (in^2) to length (in).
If possible, discuss this situatio with your teacher.
The correct answer is b because the number increases by 4 every time
hope I helped
Answer:
2/3
Step-by-step explanation:
Rise/Run. The one taught in school.
2/3
also 69 problems? That's alot. Hang in there...
Let
x----> the length side of the hypotenuse
y----> the length side of the <span>unknown leg
we know that
x=2*y----> equation 1
applying the Pythagoras Theorem
x</span>²=y²+3²-----> equation 2
substitute equation 1 in equation 2
[2*y]²=y²+3²----> 4*y²-y²=9-----> 3*y²=9----> y²=3
y=√3 ft
x=2*y----> x=2*√3 ft
the answer is
<span>the lengths of the three sides are
</span>hypotenuse=2√3 ft
one leg=√3 ft
other leg=3 ft
