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aleksandrvk [35]
3 years ago
13

What is the slope of a line that is perpendicular to the line represented by the equation x-y=8

Mathematics
2 answers:
Evgen [1.6K]3 years ago
7 0
Rewrite the equation into the general form: y=mx+b, where m represents slope.
We have:
x-y=8
-y=-x+8
y=x-8
Here, m happens to be 1. The slope of a perpendicular line to an equation is the negative reciprocal of the number. 
For example, the negative reciprocal of 2 is -1/2 (so you 'flip' the number).
So the negative reciprocal of 1, or 1/1, is -1.
So the slope of the line perpendicular to that equation is -1.
Paraphin [41]3 years ago
5 0
Slope of line is -1 , i think
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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
GuDViN [60]

Answer:

Euler's method is a numerical method used in calculus to approximate a particular solution of a differential equation. As a numerical method, we have to apply the same procedure many times, until get the desired result.

In first place, we need to know all the values the problem is giving:

  • The step size is 0.2; h = 0.2. This step size is a periodical increase of the x-variable, which will allow us to calculate each y-value to each x.
  • The problem is asking the solution y(1), which means that we have to find the y-value assigned for x = 1, through the numerical method.
  • The initial condition is y(0) = 9. In other words, x_{o} = 0\\y_{0}=9.

So, if the initial x-value is 0, and the step size is 0.2, the following x-value would be: x_{1}=0.2; then x_{2}=0.4; x_{3} =0.6; x_{4} =0.8;x_{5} =1; and so on.

Now, we have to apply the formula to find each y-value until get the match of x_{5}=1, because the problem asks the solution y(1).

According to the Euler's method:

y_{1} =y_{0} +hF(x_{0};y_{0})\\y_{2} =y_{1} +hF(x_{1};y_{1})\\y_{n} =y_{n-1} +hF(x_{n-1};y_{n-1})

Where F(x;y)=x^{2} y-12y^{2}, and x_{0} =0; y_{0} =9; h=0.2.

Replacing all values we calculate the y-value assigned to x_{1}:

y_{1} =9+0.2((0)^{2} 9-12(9)^{2})=-185.4.

Now, y_{1} =-185.4, x_{1} =0.2; h=0.2. We repeat the process with the new values:

y_{2} =y_{1} +hF(x_{1};y_{1})  \\y_{2} = -185.4+0.2((0.2)^{2} (-185.4)-12(-185.4)^{2} )\\y_{2}=-82682.47

Then, we repeat the same process until get the y-value for x_{5} =1, which is y_{5} = -1.0018, round to four decimal places.

Therefore, y(1)=-1.0018.

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AlexFokin [52]
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Answer:

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Answer:

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I have had this exact question..I got it right with 4. =-=

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