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3 years ago

What is the approximate distance between points A and B?

Mathematics

2 answers:

shepuryov [24]3 years ago

7 0

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[1-(-2)]^2+[4-(-3)]^2}\implies d=\sqrt{(1+2)^2+(4+3)^2} \\\\\\ d=\sqrt{9+49}\implies d=\sqrt{58}\implies d\approx 7.62$

Licemer1 [7]3 years ago

5 0

Answer:

7.62 units

Step-by-step explanation:

We are given that

The coordinates of point A is at (1,4).

The coordinates of point B is at (-2,-3).

We have to find the approximate distance between A and B.

Distance formula: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using this formula we will find the distance between A and B.

AB $=\sqrt{(-2-1)^2+(-3-4)^2}$

Distance between A and B= $\sqrt{49+9}=\sqrt{58}$

Distance between A and B= $7.62$ units

Hence, the approximate distance between points A and B=7.62 units

Option C is true.

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For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$

Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.

We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,

$3 = 15-6*2$

Because 4 multiplies 3 to give 12, we multiply by 4

$12 = 15*4-6*8$

So one solution is

$x=-8$ & $y = 4$

All other solutions will have the form

$x=-8+\frac{15r}{3} = -8+5r\\y=4-\frac{6r}{3} =4-2r$

where $r$ ∈ Ζ

Hence by putting r values, we get many (x, y)

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