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jeyben [28]
3 years ago
10

Please help me............... Can someone be by friends?

Mathematics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

Can someone be my friend is what I think you mean.

Yes. I could be!

<h3>Keep </h3><h3>Calm</h3><h3>You</h3><h3>Just</h3><h3>Got</h3><h3>LOKI'D!</h3><h3>EHEHEHE</h3>
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Hi how can I graph this if it dosn't have a slope?<br> 1. y = (2/3)x - 1 <br> y = -x + 4
12345 [234]

System of Linear Equations entered :

[1] y - 2x/3 = -1

[2] y + x = 4

// To remove fractions, multiply equations by their respective LCD

Multiply equation [1] by 3

// Equations now take the shape:

[1] 3y - 2x = -3

[2] y + x = 4

Graphic Representation of the Equations :

-2x + 3y = -3 x + y = 4

Solve by Substitution :

// Solve equation [2] for the variable x

[2] x = -y + 4

// Plug this in for variable x in equation [1]

[1] 3y - 2•(-y +4) = -3

[1] 5y = 5

// Solve equation [1] for the variable y

[1] 5y = 5

[1] y = 1

// By now we know this much :

y = 1

x = -y+4

// Use the y value to solve for x

x = -(1)+4 = 3

I hope this help you

8 0
3 years ago
Which box-and-whisker plot represents this data
Lady_Fox [76]
The fist box and whisker plot correctly represents the data
5 0
4 years ago
What is the solution of the question 2(x+2)^2-4=28
Arisa [49]
2((x+2)(x+2)}-4=28
2{x^2 + 4x + 4}-4=28
2x^2 +4x=28
2(x^2+2x)=28
x^2+2x=28/2
x^2+2x=14
5 0
3 years ago
3. Using techniques from Calculus, show directly that the maximum value of a 1-D Gaussian distribution occurs at the point x = μ
Vilka [71]

Answer:

For a scaler variable, the Gaussian distribution has a probability density function of

p(x |µ, σ² ) = N(x; µ, σ² ) = 1 / 2π×e^{\frac{-(x-u)^{2}}{2s^{2} }  }

The term will have a maximum value at the top of the slope of the 1-D Gaussian distribution curve that is when exp(0) =1 or when x = µ

Step-by-step explanation:

Gaussian distributions have similar shape, with the mean controlling the location and the variance controls the dispersion  

From the graph of the probability distribution function it is seen that the the peak is the point at which the slope = 0, where µ = 0 and σ² = 1 then solution for the peak = exponential function = 0 or x = µ

5 0
3 years ago
Numbers like 10, 100, 1000 are called?
Y_Kistochka [10]
Those number are all powers of 10
10= 10^1
100= 10^2
1000= 10^3
and so on...
5 0
4 years ago
Read 2 more answers
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