Sammy is classifying quadrilateral EFGH. How can she use coordinate geometry to determine whether EFGH is a rhombus? E(2, 5), F(
1 answer:
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Step-by-step explanation:
This is a probability related problem.
Probability is the likelihood of an event to occur;
Pr =
The sample space here is from 1 to 25 which is 25
A.
Pr of a card marked 8; we have just 1 possible outcome;
Pr(8) =
B.
Pr of drawing a card that is a multiple of 5;
Multiples of 5 = 5, 10, 15 and 25
Pr (multiples of 5) =
C.
Pr of drawing a card with odd numbers:
Number of odd numbers between 1 and 25 = 13
Pr(odd numbers) =
D.
Pr of drawing a number with square number on it;
Square numbers between 1 and 25 = 1, 4, 9, 16 and 25
Pr(square numbers) = =
<h3>Answer: 2.2 units</h3>
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Explanation:
I'll define these point labels
- B = Blake's starting position
- F = finish line
- C = the third unmarked point of the triangle
The locations of the points are
- B = (-8,1)
- C = (-6,-3)
- F = (4,-2)
Use the distance formula to find the distance from B to C
Segment BC is roughly 4.47214 units long.
Following similar steps, you should find that segment CF is approximately 10.04988 units long.
If Blake doesn't take the shortcut, then he travels approximately BC+CF = 4.47214+10.04988 = 14.52202 units. This is the path from B to C to F in that order.
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Use the distance formula again to find the distance from B to F. This distance is about 12.36932 units. He travels this amount if he takes the shortcut.
Subtract this and the previous result we got
14.52202 - 12.36932 = 2.1527
That rounds to 2.2
This is the amount of distance he doesn't have to travel when he takes the shortcut.
In other words, the track is roughly 2.2 units shorter when taking the shortcut.
Side note: Replace "units" with whatever units you're working with (eg: feet or meters).