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exis [7]
3 years ago
6

Please help! These mathematics are very confusing. need help right away.

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
5 0
<span>
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.
____________________________________

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) .
____________________________________________

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>
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By looking at the picture with all the possible cases, we can tell that the correct option is D.

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3 years ago
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What is the result of isolating x^2 in the equation below 10x^2-36y^2=100
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For the answer to the question above,
<span>10x^2-36y^2=100
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</span>
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2 years ago
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Find the value of x. Picture below
Wittaler [7]

Here a right angled triangle given. We know that one angle of a right angled triangle is 90°.

As the sum of three angles of a triangle is 180°, so we can say the sum of other two angles of a right angled triangle is (180-90)° = 90°.

Here in the figure the other two angles given (2x)^o and (x+15)^o. Sum of these two angles is 90°.

So we can write the equation as,

(2x)+(x+15) = 90

We have to remove the parenthesis now.

2x+x+15 = 90

Now we will add the like terms. Here x and 2x are like terms. By adding them we will get,

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To solve it for x, now we have to move 15 to the other side by subtracting it from both sides.

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Now to get x, we have to move 3 to the other side, by dividing it to both sides.

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