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2 years ago

13

xy=22 as a proportion please :) It would mean a lot if I could get a fast response! Please explain in detail because I don't get

what it is asking.

2 answers:

fgiga [73]2 years ago

7 0

Xy = 22
1 1

Cross multiply:
xy * 1 = 1 * 22

Simplifying
xy * 1 = 1 * 22

Reorder the terms for easier multiplication:
1xy = 1 * 22

Multiply 1 * 22
1xy = 22

Solving
1xy = 22

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Divide each side by '1y'.
x = 22y-1

Simplifying
x = 22y-1

Fiesta28 [93]2 years ago

3 0

The answer is 22
Have a nice day

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Read 2 more answers

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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Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

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d)Q(9,10) is true

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

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

Then

$(-5)^{2}$

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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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