A. Surveying the students about their favorite type of math problem.
and
C. Sampling water for contaminants
Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards
the distance between them is 476 miles, and they met 8.5hrs later
let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs
we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2
thus

solve for "r", to see how fast the cyclist was going
what about the car? well, the car's rate is r + 33.2
There are several ways, but the general format follows f(x) = ax2 + bx + c f(x) = ax² + bx + c, where A, B, and C are non-zero numbers. Another way of finding a quadratic equation is examining the graph of it, you'll notice a "U" shape called a parabola, which come in many shapes but they all retain a "U"-like curve.
Answer:
N=115
Step-by-step explanation:
n=.25(2.5715/.12)^2 =114.8 so round up to 115. Hope that help you
We have that
<span> -3r+3u
</span>
Step 1
<span>Use the distance formula to find the radius
</span><span>r = √(x^2 + y^2)
(x,y)----------> (</span>-3,3)
r= √(-3^2 + 3^2)=√18=3√2
Step 2
Use the tangent to find the angle
θ = arctan(y/x)
θ = arcTan(3/-3)= -45°-------> II Quadrant
then
θ=180-45=135°
Step 3
write in polar form
(x,y)------------> (r, θ)
so
(-3,3)----------> (3√2, 135°)
the answer is (3√2, 135°)