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3 years ago

5

It's geometry, not really sure what to do

1 answer:

Grace [21]3 years ago

5 0

Answer:

K'= (-1,-1)

J'= (-1,-5)

L'= (0,-3)

Step-by-step explanation:

What you do here is, input the (x,y) coordinates into the translation.

For example, the original point K is (-3,5). Insert this into the translation.

(-3,5) → (-3+2, 5-8) = (-1,-3)

Repeat this for the next coordinates of L and J.

J= (-3,3)

(-3,3) → (-3+2, 3-8) = (-1,-5)

L= (-2, 5)

(-2, 5) → (-2+2, 5-8) = (0,-3)

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Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

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a) Q(-2,1) is false

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Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

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3 years ago