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iVinArrow [24]
3 years ago
12

3 1/4 minus 1 7/8 equals

Mathematics
2 answers:
Vladimir [108]3 years ago
8 0

Answer:

3 1/4= 13/4 and 1 7/8= 15/8 and find least common factor which is 8 so 26/8-15/8=11/8 or 1 3/8

Step-by-step explanation:

Leni [432]3 years ago
6 0

echo2155, STOP DELETING MY ANSWERS

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The average speed of a car is three times as fast as the average speed of a cyclist. to travel 225 km, the cyclist requires 5 h
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Average speed of cyclist = x
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average speed of cyclist = 45km/h
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The sum of 10 and a product of 7 and a number
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Hopefully this helps!
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Read 2 more answers
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
prisoha [69]
A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
7 0
3 years ago
Answer and explanation plz ???
Nutka1998 [239]

Answer:

25

Step-by-step explanation:

Formula

2xy + 1

Givens

x = 3

y = 4

Solve

2*(3)*(4) + 1

24 + 1

25

7 0
3 years ago
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