Volume of pyramid = 1/3*AH, where A is the area of the base and H is the height
Since we are given three sides of the triangle (the base of the pyramid), we should use Heron formula: Area of a triangle = sq.root of(s(s - a)(s - b)(s - c)), where s is the semi-perimeter (ie. the perimeter divided by 2) and a, b and c are side lengths.
s = (5 + 12 + 13)/2
= 30/2 = 15
A = sq,root of(15(15 - 5)(15 - 12)(15 - 13))
= sq.root of (15*10*3*2)
= sq.root of 900
= 30 m^2
Volume = 1/3*AH
= 1/3*30*9
= 90 m^3
Answer:
2000 _< t
think ;-;...................................
hey mate,
![x^2 + 8x + 16 = 0](https://tex.z-dn.net/?f=x%5E2%20%2B%208x%20%2B%2016%20%3D%200)
<=>
![(x+4)^2 = 0](https://tex.z-dn.net/?f=%28x%2B4%29%5E2%20%3D%200)
As we know that ![(a+b)^2 = a^2+b^2+2ab](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2%20%3D%20a%5E2%2Bb%5E2%2B2ab)
then
![(x+4)^2 = 0](https://tex.z-dn.net/?f=%28x%2B4%29%5E2%20%3D%200)
<=> (x+4) = 0
<=> x = -4
so there is one solution x = -4
thanks
Least value = 56
Q1 = 73
median = 82
Q3 = 86
greatest value = 95
IQR = 13
Range = 39
Answer:
Vertex form:
The vertex is ![(-\frac{1}{2},-\frac{29}{4})](https://tex.z-dn.net/?f=%28-%5Cfrac%7B1%7D%7B2%7D%2C-%5Cfrac%7B29%7D%7B4%7D%29)
Step-by-step explanation:
For a general quadratic function the form is:
For the function
The values of the coefficients for the function are the following:
,
,
Take the value of b and divide it by 2. Then, the result obtained squares it.
Add and subtract
Write the expression of the form
The vertex is (h, k)
The vertex is: ![(-\frac{1}{2},-\frac{29}{4})](https://tex.z-dn.net/?f=%28-%5Cfrac%7B1%7D%7B2%7D%2C-%5Cfrac%7B29%7D%7B4%7D%29)