Answer:
Step-by-step explanation:
we know that
In a parallelogram the diagonals bisect each other
The figure XYZW is a parallelogram
so
VW=VY
VX=VZ
<em>Find the value of b</em>
VW=VY
substitute the given values
solve for b
Find the length of YW
----> by addition segment postulate
substitute the value of b
I don't know the official answer but I can confirm for you that it is either C or D. I just don't remember how to determine which way the sign goes. Sorry
Answer:
14.4 lb
Step-by-step explanation:
In a see-saw in equilibrium, the torque generated by one side needs to be the same generated in the other side. The torque is calculated by the product between the mass and the distance to the center of the see-saw.
The torque generated by the child is:
T1 = 60 * 3 = 180 lb*feet
So, the torque generated by the weight needs to be higher than T1 in order to lift the child.
The lowest mass is calculated when the mass is in the maximum distance, that is, 12.5 feet from the center.
So, we have that:
T2 = 180 = mass * 12.5
mass = 180/12.5 = 14.4 lb
So the lowest weight is 14.4 lb
Answer: The length and width are 50 and 30 meters (or 30 and 50 meters).
Step-by-step explanation:
To solve this question, we can represent variables for the length and width in two equations.
To solve for one of the variables, you'll have to substitute one of the variables, so solve for one of them:
Now, we have a standard quadratic equation that we can factor. When factoring, you'll get this:
This tells us that the width could be either 50 or 30.
Substitute 50 into one of the equations to find the length:
2 (l) + 100 = 160
l = 30.
The length and width are 50 and 30 meters (or 30 and 50 meters).
If an expression is the difference of two squares, it will follow this format:
where a and b are integers or variables.
1)
Since the expression follows the format, it's a DoTS.
2) Since the binomial has an addition sign, it isn't a DoTS.
3)
The second term isn't an integer, so the expression isn't a DoTS.
4)
Since the expression follows the format, it's a DoTS.
Therefore, the first and fourth expressions are differences of two squares.
