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eduard
3 years ago
15

Please answer my questions!

Mathematics
1 answer:
timama [110]3 years ago
4 0

Answer:

What is the question.

Step-by-step explanation:

Cant help u without one.

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You start at (-3, 0). You move right 3 units and down 5 units. Where do you end?
Evgen [1.6K]

Answer:

(0,-5)

Step-by-step explanation:

5 0
3 years ago
What is the exponential notation of 100
Andreyy89
The answer is 1x10^2 or just simply 10^2.
5 0
3 years ago
Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2
exis [7]

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}

with determinant |J| = 12, hence the area element becomes

dA = dx\,dy = 12 \, du\,dv

Then the integral becomes

\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv

where R' is the unit circle,

\dfrac{x^2}{16} + \dfrac{y^2}9 = \dfrac{(4u^2)}{16} + \dfrac{(3v)^2}9 = u^2 + v^2 = 1

so that

\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}

Then

\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}

3 0
2 years ago
The value of a weight vector is given as (w1=3, w2=-2, w0=1) for a linear model with soft threshold (sigmoid) function f(x).
dolphi86 [110]

Answer:

Please see explanation for the answer. The code is written in python and is as given below:

Step-by-step explanation:

The solution is obtained on the Python with the following code

import matplotlib.pyplot as plotter

import numpy as npy

x_s = npy.linspace(-5,5,100)  #Defining a linear sample space with boundaries as -5 to 5 and 100 as number of samples.

def sigmo(z):return 1/(1 + npy.exp(-z)) #Defining sigmoid function for the f(x).

plotter.plot(x_s, sigmo(x_s))

plotter.plot([-5,5],[.5,.5])

plotter.xlabel("z")

plotter.ylabel("sigmoid(z)")

plotter.show()

8 0
3 years ago
‼️‼️‼️‼️ pls, i need help
zepelin [54]
Yo soy muy grande cojones y muy grande Pena!!!!!!
7 0
3 years ago
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