The formula for finding the perimeter of a quadrilateral is Length + Length + Width + Width.
<h3>What is Perimeter?</h3>
- A perimeter is the path that surrounds a certain shape. To calculate the path that surrounds a quadrilateral, we need to get the sum of its four sides, both lengths and widths, lengths being the longest sides and the widths being the shortest.
- The formula used for calculating perimeter is Perimeter = Length + Length + Width + Width.
- For instance, to calculate the perimeter of a parallelogram with a side of 5 cm and one of 3 cm, we insert the numbers in their corresponding spot in the formula as such: Perimeter=5+5+3+3=16 cm or since parallelograms have 2 sets of 2 equal sides, we can use this formula Perimeter=(5×2)+(3×2)=10+6=16 cm.
- For a square on the other hand, we only need to know the length of one side because it has 4 equal sides.
Therefore, the formula for finding the perimeter of a quadrilateral is Length + Length + Width + Width.
Learn more about quadrilateral here:
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In intercept form, the plane that has these intercepts is ...
... x/(x-intercept) + y/(y-intercept) + z/(z-intercept) = 1
... x/1 + y/(-1) + z/2 = 1
... 2x -2y +z = 2 . . . . . in standard form
Answer:
B
Step-by-step explanation:
Data set D does not contain the value 128, which is the median value.
Data set C does not contain the outlier value 91.
Data set A contains value 168, which does not show up on the plot.
The only remaining choice is B.
_____
In order, the data values of set B are ...
... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152
The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)
Answer:
B
Step-by-step explanation:
80*17=1360/choice B
Answer:
The solution of system of equation is (-2,0)
Step-by-step explanation:
Given system of equation are
Equation 1 : 2x+y=(-4)
Equation 2 : y+
x=(-1)
To plot the equation of line, we need at least two points
For Equation 1 : 2x+y=(-4)
Let x=0
2x+y=(-4)
2(0)+y=(-4)
y=(-4)
Let x=1
2x+y=(-4)
2(1)+y=(-4)
y=(-6)
Therefore,
The required points for equation is (0,-4) and (1,-6)
For Equation 2 : y+x=(-1)
Let x=0
y+x=(-1)
y+(0)=(-1)
y=(-1)
Let x=2
y+x=(-1)
y+(2)=(-1)
y=(-2)
The required points for equation is (0,-1) and (2,-2)
Now, plot the graph using this points
From the graph,
The red line is equation 1 and blue line is equation 2
Since. The point of intersection is solution of system of equations
The solution of system of equation is (-2,0)
