Aldrin and Jan are standing at the edge of a huge field. At 2:00 p.m., Aldrin begins to walk along a straight path at a speed of 3 km/h. Two hours later, Jan takes a straight path at a 608 angle to Aldrin’s path, walking at 5 km/h. At what time the two friends be 13km apart?

Answers

Answer 1
Answer:

Answer: After 2:29 hours both Aldrin and Jan will be 13 km apart from each other.

Time at that instant will be : 4:00 pm + 2.29 hours= 6:29 pm

Step-by-step explanation:

According to the figure:

Ab= c = 13 Km

AC = b = distance= speed* time =5* t=5t Km

After two hours the distance covered by the Aldrin say till point D :

distance=speed* time=3km/hr* 2 hrs=6 Km

Since, we are interested in determining the time at which Aldrin and Jan both 13 Km apart. For that Aldrin Additional distance covered by the Aldrin from point D to B is :

distance=speed* time= 3km/hr* t=3t Km

CB= a = CD+DB= 6+3t

We will apply Laws of cosines:

C^2=A^2+B^2-2AB* Cos\theta

(13)^2=(6+3t)^2+(5t)^2-2(5t)(6+3t)Cos 60^o

19t^2+6t-133=0

On solving the the above equation by quadratic formula we get two values of 't'

1) t = 2.49 hours , 2) t = -2.80 hours

Since,time cannot be in negative so we will ignore the negative value of t

t = 2.49 = 2:29 hours ,(0.49 hour=0.49 × 60 min=29.4 min)

After 2.49 hours after the walking of Jan. Both Aldrin and Jan will be 13 km apart from each other.

Time at that instant will be : 4:00 pm + 2:29 hours= 6:29 hour


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Answers

the answer would me C 5(24+18)=210 210*1/2=105.

Room is 360 square feet. The width is 5/8 the length. What is the length

Answers

Since you are given the width of 5/8 of the length, I will assume thatthe figure is a rectangle. Let us represent W as the width and L as the length.So we have W = 5/8L. then we are also given the area of the rectangle at 360square feet. The area of the rectangle is represented by the formula A = LxW. 

A = LxW
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Karissa begins to solve the 1/2 (x-14)+11=1/2x-(x-4)

Answers

Answer:

Step-by-step explanation:

C)0

The safe fluoride-to-water ratio is 0.5 to 1.4 parts per million (ppm) gallons. What is the range in parts of fluoride that could be added to 500,000 gallons of water?

Answers

Answer:

250,000 to 700,000 parts

Step-by-step explanation:

For the lower ratio of 0.5, we have to get 0.5 flouride parts/1 gallon of water equivalent to (some fluoride parts)/ (500,000 gallons of water). we can multiply 500,000 by 0.5 to get 250,000 as our lower limit. Similarly, we can multiply 500,000 by 1.4 to get 700,000 as our upper limit

George went to school for 216 days in a full year. If his attendance in 60%, find the number of days on which the school was opened.

Answers

Given:
days present = 216 days
percentage of presence = 60%

Find the total number of days the school was opened.

216 days is the part of a whole. 60% is the percentage of the part to the whole.

To get the whole, we need to divide the part by its percentage

216 ÷ 60% = 360 days

The school was opened for 360 days.

Hi

216 days ⇒ 60% 
x days ⇒ 100% 

60x = 216·100
60x = 21600
x = 21600/60
x = 2160/6
x = 360 days

Answer: 360 days

If mZBCD = 51°, solve for x.
(1 4x + 4) 8
55°

Answers

Answer:

(14x+4)8=55 were x = 3.07.

Step-by-step explanation:

(14x + 4) 8 = 55

             -8     -8

      14x + 4 = 55

             - 4     -4

            14x = 55

                 14

            x = 3.07