4 in roman numerals is IV. V = 5, and the I is like taking one off the 5. If it was VI, it would be 6, like adding 1. So, IV is 4. The patient will take IV mLs.
No it’s not the square root of 32 is 5.65685424949
B should be the only correct answer.
The volume of cone is calculated by the formula
![V= \frac{1}{3} \pi R^2H](https://tex.z-dn.net/?f=V%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20R%5E2H)
, where H is a height of cone and R is a radius of the bottom.
In your case, R=3 cm and R=7 cm. Then
![V= \frac{1}{3} \pi* 3^2*7= \frac{63\pi}{3} =21\pi](https://tex.z-dn.net/?f=V%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%2A%203%5E2%2A7%3D%20%5Cfrac%7B63%5Cpi%7D%7B3%7D%20%3D21%5Cpi)
.
Answer: The volume of candy is 21π.
Answer:
The area of the rectangle is 42 units^2
Step-by-step explanation:
we know that
The area of rectangle is equal to
![A=LW](https://tex.z-dn.net/?f=A%3DLW)
In this problem
AB=DC
BC=AD
see the attached figure with letters to better understand the problem
we have that
![L=BC\\W=AB](https://tex.z-dn.net/?f=L%3DBC%5C%5CW%3DAB)
the formula to calculate the distance between two points is equal to
![d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28y2-y1%29%5E%7B2%7D%2B%28x2-x1%29%5E%7B2%7D%7D)
we have the points
![A(2,-1),B(5,2),C(12,-5),D(9,-8)](https://tex.z-dn.net/?f=A%282%2C-1%29%2CB%285%2C2%29%2CC%2812%2C-5%29%2CD%289%2C-8%29)
<em>Find out the distance BC</em>
we have
![B(5,2),C(12,-5)](https://tex.z-dn.net/?f=B%285%2C2%29%2CC%2812%2C-5%29)
substitute in the formula
![d=\sqrt{(-5-2)^{2}+(12-5)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-5-2%29%5E%7B2%7D%2B%2812-5%29%5E%7B2%7D%7D)
![d=\sqrt{(-7)^{2}+(7)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-7%29%5E%7B2%7D%2B%287%29%5E%7B2%7D%7D)
![d_B_C=\sqrt{98}\ units](https://tex.z-dn.net/?f=d_B_C%3D%5Csqrt%7B98%7D%5C%20units)
<em>Find out the distance AB</em>
we have
![A(2,-1),B(5,2)](https://tex.z-dn.net/?f=A%282%2C-1%29%2CB%285%2C2%29)
substitute in the formula
![d=\sqrt{(2+1)^{2}+(5-2)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%282%2B1%29%5E%7B2%7D%2B%285-2%29%5E%7B2%7D%7D)
![d=\sqrt{(3)^{2}+(3)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%283%29%5E%7B2%7D%2B%283%29%5E%7B2%7D%7D)
![d_A_B=\sqrt{18}\ units](https://tex.z-dn.net/?f=d_A_B%3D%5Csqrt%7B18%7D%5C%20units)
<em>Find out the area</em>
![A=(L)(W)](https://tex.z-dn.net/?f=A%3D%28L%29%28W%29)
we have
![L=d_B_C=\sqrt{98}\ units](https://tex.z-dn.net/?f=L%3Dd_B_C%3D%5Csqrt%7B98%7D%5C%20units)
![W=d_A_B=\sqrt{18}\ units](https://tex.z-dn.net/?f=W%3Dd_A_B%3D%5Csqrt%7B18%7D%5C%20units)
substitute
![A=(\sqrt{98})(\sqrt{18})=42.0\ units^2](https://tex.z-dn.net/?f=A%3D%28%5Csqrt%7B98%7D%29%28%5Csqrt%7B18%7D%29%3D42.0%5C%20units%5E2)