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4 years ago

9

Find the distance between the points (-6,7) and (0,8)

1 answer:

natka813 [3]4 years ago

8 0

We can solve this problem by using the distance formula. The distance formula is: $\sqrt{(x_{1}-x)^{2} + (y_{1} - y)^{2}}$ We can now put in values and solve.

$\sqrt{(-6-0)^{2} + (7-8)^{2}}$

$\sqrt{36 + 1}$

$\sqrt{37}$

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26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

3 0

3 years ago

List three examples of times when people need to add or subtract fractions in there daily lives.

Brut [27]

Taxes
sale discounts
cooking

8 0

3 years ago

Which is a point on the circle whose center is (0,0) and whose radius is 5? A. (0,0)

Temka [501]

Answer:

(3,4)

Step-by-step explanation:

x^2+y^2=(5)^2

if you graph it you will see that (3,4) is on the circle

4 0

3 years ago

Simplyfy the given equation 5x+2(x-3)=-2(x-1)

Blizzard [7]

5x+2x−6=−2x+2

7x−6=−2x+2

7x=−2x+2+6

7x=−2x+8

7x+2x=8

9x=8

x=8/9

Hope this helps, have a blessed day! :-)

5 0

3 years ago

Only one question pls help

lina2011 [118]

Cos x = 5 / 9 ==》 Cos x = 0.55

==》 x = 56 degrees

4 0

3 years ago