The points M(-7,2), N(-3,4), O(-5,8)M(−7,2),N(−3,4),O(−5,8), and P(-9,6)P(−9,6) form a quadrilateral. Find the desired slopes an
pochemuha
The slope of line PO and MN is 0.5. And the slope of the line ON and MP is negative 2. Then each length of the quadrilateral is 4.47 units.
<h3>What is a quadrilateral?</h3>
It is a quadrilateral with four sides. The total interior angle is 360 degrees.
The points M(−7,2), N(−3,4), O(−5,8), and P(−9,6) form a quadrilateral.
Then the slope of each line of the quadrilateral will be
The slope of the line PO will be
PO = (8-6)/(-5+9)
PO = 2/4
PO = 0.5
The slope of the line ON will be
ON = (8-4)/(-5+3)
ON = 4/(-2)
ON = -2
The slope of the line MN will be
MN = (4-2)/(-3+7)
MN = 2/4
MN = 0.5
The slope of the line MP will be
MP = (6-2)/(-9+7)
MP = 4/(-2)
MP = -2
Then each length of the quadrilateral will be
The length of PO will be
PO = √[(-9+5)² + (6-8)²]
PO = 4.47
The length of ON will be
ON = √[(-5+3)² + (8-4)²]
ON = 4.47
The length of MN will be
MN = √[(-3+7)² + (4-2)²]
MN = 4.47
The length of MP will be
MP = √[(-9+7)² + (6-2)²]
MP = 4.47
More about the quadrilateral link is given below.
brainly.com/question/13805601
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Treat it like a fraction
18b^2/45b
divide top and bottom by b
18b/45
divide top and bottom by 3
6b/15
divide top and bottom by 3 again
2b/5
2b:5 is the simpliest form of the ratio
Answer:
Step-by-step explanation:
A) 9.56x10^38 ergs B) 7.4x10^-3 mm A). For the sun, just multiply the power by time, so 3.9x10^33 erg/sec * 2.45x10^5 sec = 9.56x10^38 B) Of the two values 7.4x10^-3 and 7.4x10^3, the value 7.4x10^-3 is far more reasonable as a measurement for blood cell. Reason becomes quite evident if you take the 7.4x10^3 value and convert to a non-scientific notation value. Since the exponent is positive, shift the decimal point to the right. So 7.4x10^3 mm = 7400 mm, or in easier to understand terms, over 7 meters. That is way too large for a blood cell when you consider that you need a microscope to see one. Now the 7.4x10^-3 mm value converts to 0.0074 mm which is quite small and would a reasonable size for a blood cell.
