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Varvara68 [4.7K]
2 years ago
14

25 POINTS! Determine the slope in the graph and be able to explain how you found it.

Mathematics
2 answers:
Lelechka [254]2 years ago
7 0
12 points aight, my fault
DiKsa [7]2 years ago
6 0

Answer:

slope = 5

Step-by-step explanation:

For a straight line, slope  = change in vertcal coordinate/change in horizontal component. Students sometimes remember this as "rise over run", because it measures how much the graph rises (up for positive or down for negative) for every unit it 'runs' to the right.

Pick any two different points from the graph, for example (x,y) = (24, 120) and (x,y) = (4, 20).

The change in y = 120-20 = 100

The change in x = 24-4 = 20

(It doesn't matter which point's coordinates you put first in the subtractions, as long as it is the <u>same</u> point in both the numerator and the denominator.)

slope = (change in y)/(change in x) = (120-20)/(24-4) = 100/20 = 5

If you first look at the graphic in this problem, you might think the slope is 1, because the line goes up 1 square for every square you go to the right.  But that is because of the scale on the two axes. The x-axis coordinate only increases by 4 units for each block to the right that you move. But the the y-axis coordinate increases by 20 units for each block up that you move. Thus, a block on the y-axis is equivalent to 5 blocks on the x-axis!

In math terminology, you may see slope written as m = Δy/Δx, because the usual variable for slope is 'm' and the Greek letter delat 'Δ' is sued in math, scince, and enginerring as a symbol for 'change'.

Hope this is a good explanation for you.

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Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3√5 , KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMN
lora16 [44]
Q1)Find KM
As ME is perpendicular to KN, ∠KEM is a right angle
Therefore ΔKEM is a right angled triangle 
KE is given and and ME is also given, we need to find KM
for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
KM² = 8² + (3√5)²
       = 64 + 9x5
KM = √109
KM = 10.44

Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
and the length of KN is 5x
KN is greater than LM by 2 units 
If we take the figure ∠K and ∠N are equal. 
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units 
KN = LM + 2x 
2x = KE + DN
2x = 8+8
x = 8
LM = 3x = 3*8 = 24

Q3)Find KN 
Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have;
same height ME = LD perpendicular distance between the 2 parallel sides 
same right angle when the perpendicular lines cut KN
∠K = ∠N 
when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN 
the distance ED = LM
Therefore KN = KE + ED + DN
 since ED = LM = 24
and KE + DN = 16
KN = 16 + 24 = 40
another way is since KN = 5x and x = 8
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Q4)Find area KLMN
Area of trapezium can be calculated using the following general equation 
Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)
where perpendicular height - ME
2 parallel sides are KN and LM
substituting values into the general equation
Area = 1/2 * ME * (KN+ LM) 
         = 1/2 * 3√5 * (40 + 24)
         = 1/2 * 3√5 * 64
         = 3 x 2.23 * 32
         = 214.66 units²


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