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Exercise #1:
Point H = (–2, 2)
Point J = (–2, –3)
Point K = (3, –3)
It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it. Then
you'd immediately see what's going on.
Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
</span><span>Notice that points J and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.
Notice that point-J is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-J is one corner of the square.
H is another corner of the square. It's 5 units above J.
K is another corner of the square. It's 5 units to the right of J.
The fourth corner is (2, 3) ... 5 to the right of H,
and 5 above K.
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Exercise #2:
</span><span>Point H = (6, 2)
Point J = (–2, –4)
Point K = (-2, y) .
</span><span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it. Then
you'd immediately see what's going on.
</span><span>Notice that points J and K have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
We need K to connect to point-H in such a way that it's on
the same horizontal line as H. Then the vertical and horizontal
lines that meet at K will be perpendicular, and we'll have the
right angle that we need there to make the right triangle.
So K and H need to have the same y-coordinate.
H is the point (6, 2). So K has to be up at (2, 2) .
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Exercise #3:
</span>
<span>Point H = (-6, 2)
Point J = (–6, –1)
Point K = (4, 2) .
</span>
<span>It would be very helpful if you could take a pencil and a piece
of paper, and sketch a graph with these points on it. Then
you'd immediately see what's going on.
This exercise is exactly the same as #1, except that it's a
rectangle instead of a square. It's still make of horizontal
and vertical lines, and that's all we need to know in order
to solve it.</span><span>
Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
</span><span>Notice that points H and K have different x-coordinates but
the same y-coordinate, so they're on the same horizontal line.
Notice that point-H is on both the horizontal line and the vertical
line, so the lines meet there, and they're perpendicular.
Point-H is one corner of the rectangle.
J is another corner of the rectangle. It's 3 units below H.
K is another corner of the square. It's 4 units to the right of H.
The fourth corner is (2, -1) ... 4 to the right of J,
and 3 below K.
</span>
Answer:
2, √72, 8.7, √83, 9.25
Step-by-step explanation:
First, you need to find the integers of each number that is not already a integer.
Square root of 72: 8.48528137424
9.25
Square root of 83: 9.11043357914
2
8.7
Now, list the numbers by their whole number (in the ones place). If they have the same whole number, look at their tenths place and see which one is greater.
Overall, the answer would be 2, √72, 8.7, √83, 9.25.
Answer:
Step-by-step explanation:
y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)
so the equation become
2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0
when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.
