Answer:
12
Step-by-step explanation:
Area of a triangle = (1/2) * base * height
base width = 4 squares
height = 6 squares
area = (1/2) * 4 * 6 = 12
Answer:
What figure?
Step-by-step explanation:
I can help, just need figure :)
You can post it and ill edit this answer
Answer:
x=-5/9
Step-by-step explanation:
Answer:
85 cm²
Step-by-step explanation:
The base is a square with side length 5 cm. The area of this base is
A = s² = (5 cm)² = 25 cm².
If the slant height of each lateral face of the pyramid is 6 cm, then we can find the area of each such face using the area-of-a-triangle formula,
A = (1/2)(base)(height), which here is
A = (1/2)(5 cm)(6 cm) = 15 cm² per lateral side.
The total area of these sides is 4(15 cm²) = 60 cm².
Then the total surface area of the pyramid is A = 60 cm² + 25 cm² = 85 cm².
Answer:
The laplace transform is ![\frac{2}{(s+6)^3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28s%2B6%29%5E3%7D)
Step-by-step explanation:
We will solve this problem by applying the laplace transform properties (their proofs are beyond the scope of this explanation).
Consider first the function f(t) = 1. By definition of the laplace transform, we have
![F(s) = \int_{0}^{\infty}f(t)e^{-st}dt](https://tex.z-dn.net/?f=F%28s%29%20%3D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%7Df%28t%29e%5E%7B-st%7Ddt)
when f(t) = 1 we get
![F(s) = \int_{0}^{\infty}e^{-st}dt = \left.\frac{-e^{-st}}{s}\right|_{0}^{\infty} = \frac{1}{s}](https://tex.z-dn.net/?f=F%28s%29%20%3D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%7De%5E%7B-st%7Ddt%20%3D%20%5Cleft.%5Cfrac%7B-e%5E%7B-st%7D%7D%7Bs%7D%5Cright%7C_%7B0%7D%5E%7B%5Cinfty%7D%20%3D%20%5Cfrac%7B1%7D%7Bs%7D)
We will apply the following properties: Define L(f) as applying the laplace transform
(this means, multiplying by an exponential corresponds to a shift in the s parameter of the transform of f)
(this is, multypling by
is equivalent to taking the n-th derivative of the transform.
We are given the function
. Since the transform of the constant function 1 is 1/s, by applying the first property we get
![L(e^{-6t}\cdot 1 ) = \frac{1}{s+6}](https://tex.z-dn.net/?f=L%28e%5E%7B-6t%7D%5Ccdot%201%20%29%20%3D%20%5Cfrac%7B1%7D%7Bs%2B6%7D)
By applying the second property we get
![L(g(t)) = L(t^2 e^{-6t} \cdot 1) = (-1)^2\frac{d^2}{ds^2}(\frac{1}{s+6}) = \frac{2}{(s+6)^3}](https://tex.z-dn.net/?f=L%28g%28t%29%29%20%3D%20L%28t%5E2%20e%5E%7B-6t%7D%20%5Ccdot%201%29%20%3D%20%28-1%29%5E2%5Cfrac%7Bd%5E2%7D%7Bds%5E2%7D%28%5Cfrac%7B1%7D%7Bs%2B6%7D%29%20%3D%20%5Cfrac%7B2%7D%7B%28s%2B6%29%5E3%7D)