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Debora [2.8K]
2 years ago
15

A caravan crossed 1,378 miles of desert in 85 days. it traveled 22 miles on the first day and 28 miles on the second day. if the

caravan traveled the same number of miles on each of the remaining days, how many miles did it travel on each of those day?
Mathematics
2 answers:
vlabodo [156]2 years ago
8 0
Um idk I forgot how to do this but ima ask my brother for you so he can tell you ok
Angelina_Jolie [31]2 years ago
8 0

Answer:

The caravan traveled 16 miles each day

Step-by-step explanation:

The caravan covers a distance of 1378 miles in 85 days. The first day he covered a distance of 22 miles and the second day it covered a distance of 28 miles. The remaining days will be 85 days minus the 2 days we already know the distance the caravan covered which is 83 days.

According to the question the remaining days he covered has equal distance for each day.

let y = miles covered each day in the remaining 83 days  

So total distance covered in those 83 days is 83 × y

Total distance covered for 85 days = 1378 miles

first day  = 22 miles

second day = 28 miles

Remaining 83 days = 83 × y

1378 = 22 + 28 + 83 × y

1378 - 22- 28 = 83y

1328 = 83y

y = 1328/83

y = 16

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Step-by-step explanation:

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1 year ago
Annie is creating a stencil for her artwork using a coordinate plane. The beginning of the left edge of the stencil falls at (2,
taurus [48]

Answer:

(A)(12, 9)

Step-by-step explanation:

Given:

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A point, say Q on the stencil is at  (4, 1).

Point Q divides the stencil into the ratio 1:4.

We are required to find the end of the stencil.

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For internal division of a line with beginning point (x_1,y_1) and end point (x_2,y_2) in the ratio m:n, we use the formula

Q(x,y)=(\dfrac{mx_2+nx_1}{m+n} ,\dfrac{my_2+ny_1}{m+n} )

(x_1,y_1)=(2, -1), (x_2,y_2)=?, Q(x,y)=(4,1), m:n=1:4

Therefore:

(4,1)=(\dfrac{1x_2+4*2}{1+4} ,\dfrac{1y_2+4*-1}{1+4} )\\(4,1)=(\dfrac{x_2+8}{5} ,\dfrac{y_2-4}{5} )\\$Therefore:\\\dfrac{x_2+8}{5}=4\\x_2+8=4X5\\x_2=20-8=12\\$Similarly\\\dfrac{y_2-4}{5}=1\\y_2-4=5\\y_2=4+5=9\\(x_2,y_2)=(12,9)

The correct option is A.

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2 years ago
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Step-by-step explanation:

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2 years ago
There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

6 0
2 years ago
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