<span>ABCD is a parallelogram.
Looking at the quadrilateral ABCD, the first thing to do is to determine if the opposite sides are parallel to each other. So let's check that by looking at the opposite sides.
Line segment BA. When you go from point B to point A, you move to the right 1 space, and down 4 spaces. So the slope is -4. Looking at line segment CD, you also move to the right 1 space and down 4 spaces, which also means a slope of -4. So those two sides are parallel. When you compare line segments BC and AD, you'll notice that for both of them, you go to the right 5 spaces and up 2 spaces, so those too are parallel. So we can now saw that the quadrilateral ABCD is a parallelogram.
Since ABCD is a parallelogram, we now need to check if it's a rectangle (we know it can't be a square since the sides aren't all the same length). An easy way to test if it's a rectangle is to check of one of the angles is 90 degrees. And if we draw a line from B to D, we can create a triangle ABD. And in a right triangle, due to Pythagora's theorem we know that A^2 + B^2 = C^2 where A is the line segment AB, B is the line segment AD and C is the line segment BD. So let's calculate A^2, B^2, and C^2.
A^2: Line segment AB. We can construct a right triangle with A = 1 and B = 4. So C^2 = 1^2 + 4^2 = 1 + 16 = 17. So we have an A^2 value of 17
B^2: Line segment AD. We can construct a right triangle with A = 2 and B = 5. So C^2 = 2^2 + 5^2 = 4 + 25 = 29. So we have an B^2 value of 29
C^2: Line segment BD. We can construct a right triangle with A = 2 and B = 6. So C^2 = 2^2 + 6^2 = 4 + 36 = 40. So we have a C^2 value of 40.
Now let's check if the equation A^2 + B^2 = C^2 is correct:
17 + 29 = 40
46 = 40
And since 46 isn't equal to 40, that means that ABCD can not be a rectangle. So it's just a parallelogram.</span>
The two numbers that add to make twenty and have a difference of 4 are 12 and 8
Answer:
4
Step-by-step explanation:
Class width is said to be the difference between the upper class limit and the lower class limit consecutive classes of a grouped data. To calculate class width, this formula can be used:
CW = UCL - LCL
Where,
CW= Class width
UCL= Upper class limit
LCL= Lower class limit
From the table above:
For class 1, CW = 64 - 60 = 4
For class 2, CW = 69 - 65 = 4
For class 3, CW = 74 - 70 = 4
For class 4, CW = 79 - 75 = 4
For class 5, CW = 84 - 80 = 4
Therefore, the class width of the grouped data = 4
You're trying to find constants
such that
. Equivalently, you're looking for the least-square solution to the following matrix equation.
To solve
, multiply both sides by the transpose of
, which introduces an invertible square matrix on the LHS.
Computing this, you'd find that
which means the first choice is correct.
Rises up 3 and run is 2 so your answer is 3/2
