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zhuklara [117]
3 years ago
12

The Oxy coordinate plane for two parallel lines a and a' has the equations 2x - 3y-1 = 0 and 2x - 3y + 5 = 0. respectively. Whic

h vector translation to convert a to a'
Mathematics
1 answer:
Andru [333]3 years ago
7 0

Answer:

Remember that a vector translation can be written as:

T(a, b)

And if we apply this to a random point, (x, y), the translation gives:

T(a, b)(x, y) = (x + a, y + b)

now, remember that a general line can be written as:

y = m*x + s

Then a point of that line can be written as: (x, m*x + s)

Then if we apply the translation to a point in the line, we get:

T(a, b)(x, m*x + s) = (x + a, m*x + s + b)

Here we have two lines:

2x - 3y - 1 = 0

2x - 3y + 5 = 0

First, let's rewrite both of these in the slope-intercept form:

y = (2/3)*x - 1/3

y = (2/3)*x + 5/3

Now let's assume that we apply a translation to the first line, that has points of the form (x,  (2/3)*x - 1/3), such that we want to get points of the form:

(x, (2/3)*x + 5/3).

Then we must have:

T(a, b)(x,  (2/3)*x - 1/3) = (x + a,  (2/3)*x - 1/3 + b) = (x, (2/3)*x + 5/3).

Then we need to solve:

(x + a,  (2/3)*x - 1/3 + b) = (x, (2/3)*x + 5/3).

This means that:

x + a = x

(2/3)*x - 1/3 + b = (2/3)*x + 5/3

From the first equation, we can see that a = 0

Now we can solve the second one to find the value of b.

(2/3)*x - 1/3 + b = (2/3)*x + 5/3

subtracting (2/3)*x in both sides, we get:

-1/3 + b = 5/3

b = 5/3 + 1/3

b = 6/3 = 2

b = 2

Then the vector translation is:

T(0, 2)

So it moves the whole line 2 units upwards.

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amm1812
The answer would be 9.7
4 0
3 years ago
How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?
MAVERICK [17]

Non-negative integers are positive integers or zero.

1. When x_1=0, then there are such possible cases for x_2 and x_3:

  • x_2=0,\ x_3=10;
  • x_2=1,\ x_3=9;
  • x_2=2,\ x_3=8;
  • x_2=3,\ x_3=7;
  • x_2=4,\ x_3=6;
  • x_2=5,\ x_3=5;
  • x_2=6,\ x_3=4;
  • x_2=7,\ x_3=3;
  • x_2=8,\ x_3=2;
  • x_2=9,\ x_3=1;
  • x_2=10,\ x_3=0.

In total 11 solutions for x_1=0.

2. For x_1=1, there are such possible cases for x_2 and x_3:

  • x_2=0,\ x_3=9;
  • x_2=1,\ x_3=8;
  • x_2=2,\ x_3=7;
  • x_2=3,\ x_3=6;
  • x_2=4,\ x_3=5;
  • x_2=5,\ x_3=4;
  • x_2=6,\ x_3=3;
  • x_2=7,\ x_3=2;
  • x_2=8,\ x_3=1;
  • x_2=9,\ x_3=0.

In total 10 solutions for x_1=1.

3. This process gives you

  • for x_1=2 - 9 solutions;
  • for x_1=3 - 8 solutions;
  • for x_1=4 - 7 solutions;
  • for x_1=5 - 6 solutions;
  • for x_1=6 - 5 solutions;
  • for x_1=7 - 4 solutions;
  • for x_1=8 - 3 solutions;
  • for x_1=9 - 2 solutions;
  • for x_1=10 - 1 solution.

4. Add all numbers of solutions:

11+10+9+8+7+6+5+4+3+2+1=66.

Answer: there are 66 possible solutions (with non-negative integer variables)

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