The circle goes thru the given points <span>(3,-4) And (3,5). Since the x-coordinate does not change, the diameter lies right on the vertical line x=3. Since the vertical distance between these points is 5-(-4), or 9, the radius is 9/2.
The standard eq'n of a circle is (x-h)^2 + (y-k)^2 = r^2
Let's subst. what we know: we get (x-3)^2 + (y-1/2)^2 = (9/2)^2. Note that y = 1/2 is halfway between y=-4 and y=5.</span>
A rectangle has two dimensions, width and length, and the area of it is their product.
since we know its area is 3x²-11x-4, then the two factors from that trinomial are the likely width and length, namely (3x + 1) (x - 4).
you can check them with FOIL.
(2200 rev/min) x (360 deg/rev) x ( min/60 sec) = 13,200 degrees per second
(2200 rev/min) x (2 pi radians/rev) x (min / 60 sec) = <u>73 and 1/3 pi</u> radians/sec
Start by decomposing the number inside the root into primes
Then group the terms into cubes if possible
rewrite the root
![\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D%5Csqrt%5B3%5D%7B10%5Ccdot2%5E3%7D)
then cancel the terms that are cubes and bring them out of the root
![\sqrt[3]{80}=2\sqrt[3]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D2%5Csqrt%5B3%5D%7B10%7D)
What is the apply??? Put all of the select all that apply so I can help you
