Please help! These mathematics are very confusing. need help right away.Three vertices of a square are (–2, 2), (–2, –3), and (3, –3).
What is the fourth vertex of the square?
A.
(–3, 2)
B.
(–2, 3)
C.
(3, 2)
D.
(3, 4)
The vertices of a right triangle are (6, 2), (–2, –4), and (–2, y).
What is the value of y?
A.
–4
B.
–2
C.
2
D.
4
Three vertices of a rectangle have coordinates (–6, 2), (–6,–1), and (4, 2).
What are the coordinates of the fourth vertex of the rectangle?
A.
(–6, –2)
B.
(–6, 1)
C.
(4, –1)
D.
(4, –2)
Answers
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.
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:
Point H = (6, 2)
Point J = (–2, –4)
Point K = (-2, y) .
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 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:
Point H = (-6, 2)
Point J = (–6, –1)
Point K = (4, 2) .
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.
Notice that points H and J have the same x-coordinate, but
different y-coordinates, so they're on the same vertical line.
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.
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Answers
m = y2-y1/x2-x1
m = -6+2/2-1
m = -4/1
m = -4
Now we can plug this and one of the points into point-slope form:
y - y1 = m(x - x1)
Where y1 is the y-value of the point, x1 is the x-value of the point, and m is the slope.
y + 2 = -4(x - 1)
Distribute -4 into the parenthesis:
y + 2 = -4x + 4
Subtract 2 to both sides:
y = -4x + 2
Answers
so
make bottom 1
3/4 times 4/3=12/12=1 so
times wole thing by
result
basically
a/b divided by c/d=a/b times d/c, invert and multiply
Answers
Answers
The required sum of three consecutive number is 41.
Given that,
The product of 3 consecutive prime numbers is 2431.
We have to find,
What is their sum of 3 consecutive prime numbers.
According to the question,
A series of consecutive prime numbers to be a series of numbers, each prime, in which there are no other prime numbers between them.
For finding the 3consecutive numbers is 2431.
Taking LCM of 2431.
The LCM of 2431 is 11×13×17
The product these three numbers is 11×13×17 = 2431
And Sum of these three numbers is = 11 + 13 + 17 = 41.
Hence, The required sum of three consecutive number is 41.
For more information about Consecutive Numbers click the link given below.
Answers