2/3 rational or irrational
1 answer:
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Answer:
is outside the circle of radius of
centered at
.
Step-by-step explanation:
Let and
denote the center and the radius of this circle, respectively. Let
be a point in the plane.
Let denote the Euclidean distance between point
and point
.
In other words, if is at
while
is at
, then
.
Point would be inside this circle if
. (In other words, the distance between
and the center of this circle is smaller than the radius of this circle.)
Point would be on this circle if
. (In other words, the distance between
and the center of this circle is exactly equal to the radius of this circle.)
Point would be outside this circle if
. (In other words, the distance between
and the center of this circle exceeds the radius of this circle.)
Calculate the actual distance between and
:
.
On the other hand, notice that the radius of this circle, , is smaller than
. Therefore, point
would be outside this circle.
Solve the following system:{12 x = 54 - 6 y | (equation 1)-17 x = -6 y - 62 | (equation 2)
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1
Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1
Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.