<span>So we want to know how much clay did Joseph add after he built the cone. So the formula for the volume of the cone is V=(1/3)*pi*r^2*h where r is the radius and h is height. We know h1=12cm and r1=6cm, r2=6cm and h2=18 cm. So to get the amount of added clay Va we simply subtract the volume of the clay of the first cone V1 from the volume of the second cone V2: Vd=V2-V1=(1/3)*pi*(r1^2)*h1 - (1/3)+pi*(r2^2)*h2. Va=678.24 cm^3-452.39 cm^3= 266.08 cm^3.</span>
Answer:
The correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles
Step-by-step explanation:
In triangle ABD:
∠B = ∠D
Thus AB=AC by the property of opposite sides of equal angles are equal
In triangle CBD
∠B = ∠D
Thus CB=CD by the property of opposite sides of equal angles are equal
Thus all four sides of quadrilateral ABCD are equal
And diagonal BD bisects the angles
So, it is a rhombus
Therefore the correct option is parallelogram ABCD is a rhombus, because the diagonal bisects two angles....
Heck naw Fawk that -3-!!!!!
Answer:
question 9
= –12
while question 11=, x= –72
Step-by-step explanation:
question 9=
u= –16+4= –12
question 11=
cross multiplication
x = -36 multiplied by 2 =
-72
I HOPE THIS HELPED IF WRONG IM SORRY
and 
. The squared distance between any point in this region 
and the point (1, 1) is
![=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Br%5E2-2r%28%5Ccos%5Ctheta-%5Csin%5Ctheta%29%2B2%5Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D)
