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Alchen [17]
3 years ago
13

Cody was 165cm tall on the first day of school this year, which was 10% taller than he was on the first day of school last year.

Mathematics
1 answer:
Solnce55 [7]3 years ago
7 0
On the first day of school last year, Cody was 148.5 cm tall.
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What ere the coordinates of the point on the directed line segment from (-1,7) to (7,7) that partitions the segment into a ratio
djverab [1.8K]

Answer:

(\frac{37}{11},\frac{77}{11}  )

Step-by-step explanation:

Given two points are (-1,7) and (7,7).

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(\frac{mx2+nx1}{m+n} ,\frac{my2+ny1}{m+n} )

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(x1,y1)=(-1,7)\\\\(x2,y2)=(7,7)

also m:n=6:5 ,

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(\frac{6*7+5*-1}{11} ,\frac{6*7+5*7}{11} )

(\frac{37}{11},\frac{77}{11}  )

5 0
3 years ago
UPPER AND LOWER BOUNDS - PLEASE HELP!
Marianna [84]
Qn. 1
Lower bound for Zoe's weight = 62 - (1/2) = 62 - 0.5 = 61.5 kg

Qn. 2
Upper bound for length AB = 8.3+ (0.1/2) = 8.3+0.05 = 8.35 cm

Qn. 3
Upper bound for Anu's wight = 83+(0.5/2) = 83+0.25 = 83.25 kg

Qn. 4
Lower bound for length CD = 27-(0.5/2) = 27-0.25 = 26.75 cm

Qn. 5
Upper bound for sides of the hexagon = 3.6+(0.1/2) = 3.6+0.05 = 3.65 cm
Upper bound for the perimeter = upper bound for the sides*6 = 3.65*6 = 21.9 cm

Qn. 6
Perimeter = 4*length => side = Perimeter/4 = 24/4 = 6
Bound = 0.5/4 = 0.125
Lower bound of the length = 6-0.125 = 5.875 cm

Qn. 7
For the area,
Upper bound = 80+(10/2) 80+5 = 85 cm^2
For the length
Upper bound = 12+(1/2) = 12+0.5 = 12.5

Then, upper bound for the width = Upper bound for the area/upper bound for the length = 85/12.5 = 6.8 cm

Qn. 8
Lower bound for the area = 230-(1/2) = 230-0.5 = 229.5 cm^2
Lower bound for the sides of the square = Sqrt(Lower bound of the area) = Sqrt (229.5) = 15.15
Then,
Lower bound of perimeter = 4(Length) = 4*15.15 = 60.6 cm
8 0
3 years ago
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