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krek1111 [17]
3 years ago
12

bonnie runs 1 2/3 times as far as john each day if bonnie runs 5 miles on monday 3 1/2 miles on tuesday and 5 3/4 miles on wedne

sday how many miles did john run in all those three days
Mathematics
1 answer:
nignag [31]3 years ago
6 0

Answer:

3 miles on Monday

2¹/₁₀ miles on Tuesday

3⁹/₂₀ miles on Wednesday

Step-by-step explanation:

We are told that Bonnie runs 1⅔ (⁵/₃) times as far as John each day

This means that John runs ³/₅ times as far as Bonnie.

MONDAY:

Bonnie runs 5 miles on Monday, therefore, John runs:

³/₅ * 5 = 3 miles

TUESDAY:

Bonnie runs 3½ (⁷/₂) miles on Tuesday, therefore, John runs:

³/₅ * ⁷/₂ = ²¹/₁₀ = 2¹/₁₀ miles

WEDNESDAY:

Bonnie runs 5³/₄ (²³/₄) miles on Wednesday, therefore, John runs:

³/₅ * ²³/₄ = ⁶⁹/₂₀ = 3⁹/₂₀ miles

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The sum of two consecutive integers is 201 find the integers
telo118 [61]

100 and 101

the two integers are n and n+1. this means that 2n+1=201. n=100 and n+1=101

8 0
3 years ago
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The cross section of a water bin is shaped like a trapezoid. The bases a a trapezoid are 20 feet and 6 feet long. It has a area
nekit [7.7K]

Answer:

the height of the cross section is 3 feet

Step-by-step explanation:

The computation of the height of the cross section is shown below:

Area = 1 ÷ 2 × (a + b) × h

39 = 1 ÷ 2 × (20 + 6) × h

39 =  1 ÷2 × 26 × h

39  = 26 ÷ 2 × h

39 = 13 × h

h = 39 ÷ 13

= 3 feet

hence, the height of the cross section is 3 feet

3 0
3 years ago
Resuelve los siguientes sistemas de ecuaciones lineales utilizando el método de igualación
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Answer:

A

Step-by-step explanation:

7 0
3 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ftys along
Sever21 [200]

25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path. This can be obtained by considering this as a right angled triangle.

<h3>How fast is the tip of his shadow moving?</h3>

Let x be the length between man and the pole, y be the distance between the tip of the shadow and the pole.

Then y - x will be the length between the man and the tip of the shadow.

Since two triangles are similar, we can write

\frac{y-x}{y} =\frac{6}{15}

⇒15(y-x) = 6y

15 y - 15 x = 6y

9y = 15x

y = 15/9 x

y = 5/3 x

Differentiate both sides

dy/dt = 5/3 dx/dt

dy/dt is the speed of the tip of the shadow, dx/dt is the speed of the man.

Given that dx/dt = 5 ft/s

Thus dy/dt = (5/3)×5 ft/s

dy/dt = 25/3 ft/s

Hence 25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path.

Learn more about similar triangles here:

brainly.com/question/8691470

#SPJ4

3 0
2 years ago
PLEASE PLEASE HELP ME WITH THIS QUESTION PLEASE <br><br>AND SHOW YOUR STEPS PLEASE help me please
jeyben [28]

Answer:

4a)= 2.1m    4b)= 1.9 m    5)= 29.6º

Step-by-step explanation:

4a):

x is the opp

AB is the hyp

BC is the adj

thus,

sin32 = x/4

cross multiply, x= 4*sin32

x= 2.1196.......

x=2.1 m

4b).

x is the hyp

ac is the opp

bc is the adj

thus,

cos65=4.5/x

cross multiply, x= 4.5*cos65

x=1.9017....

x=1.9 m

5).

AB is the opp

bc is the adj

ac is the hyp

thus:

tanC= 2.1/3.7

C= tan^-1 (2.1/3.7)

C= 29.5778.....

C= 29.6º

5 0
3 years ago
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