6 + 2 < 9 + 3...u r correct
Answer:
The area of the shape is
.
Step-by-step explanation:
The shape in the graph is a composite figure is made up of several simple geometric figures such as triangles, and rectangles.
Area is the space inside of a two-dimensional shape. We can also think of area as the amount of space a shape covers.
To calculate the area of a composite shape you must divide the shape into rectangles, triangles or other shapes you can find the area of and then add the areas back together.
First separate the composite shape into three simpler shapes, in this case two rectangles and a triangle. Then find the area of each figure.
To find the area of a rectangle, we multiply the length of the rectangle by the width of the rectangle.
The area of the first rectangle is 
The area of the second rectangle is 
The area of a triangle is given by the formula
where <em>b</em> is the base and <em>h</em> is the height of the triangle.
The area of the triangle is 
Finally, add the areas of the simpler figures together to find the total area of the composite figure.

Hie!! I will be solving the questions in the sequence of the attachments.
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4. As line SU and TV are intersecting each other at O.
A.T.Q, ∠UOV = (7x - 4)°
∠SOT = 87°
Then, ∠UOV = ∠SOT {Vertically Opposite Angles}





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5. radius = 17ft
circumference of circle = 2πr


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3. ∠1 and ∠2 - add to 180 (linear pair)
∠1 and ∠3 - equal (Vertically Opposite Angles)
∠3 and ∠4 - add to 180 (linear pair)
∠2 and ∠4 - equal (Vertically Opposite Angles)
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2. 160° + (6x - 16)° = 180° (linear pair)






Answer:
1) Is more representative
Step-by-step explanation:
The problem with his selection is that maybe there are few students participating in certain sport and those students maybe do quite more excercise than the rest (or quite less). This will modify the results because the sample he selected is biased. This problem wont be solved by method 3 or 4, because he is still selecting students that may modify heavily the results with a high probability
This problem will also appear if he choose a sample by class. Maybe, in a class there are quite few students, and selecting from class will make those students appear quite more often than, lets say, a 7th grade student selected at random, therefore the selection is biased in this case as well.
If he has a list with all seventh grade students, each student is equallly likely to be selected and as a consequence, the the results wont be biased. Approach 1 is the best one.
Step-by-step explanation:
8x-5= 6x + 1
8x - 6x = 1+5
2x = 6
x= 2/6 = 1/3
x= 0.333