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# The amount of gasoline use by a car to travel varies jointly as the distance travelled and the square root of thespeed. suppose the car used 25l on a 100km trip from turod to up-uplas at 100kph. about how many litters willit use on a 192km trip at 64kph? ​

• Réponse publiée par: trizianichole20

24liters

Step-by-step explanation:

1st you will find what is the square root of 64kph and then you divide the 192km by the square root of 64kph

• Réponse publiée par: 09389706948
38.4 liter of gasoline will be used in a 192 km trip at 64 km/hr
• Réponse publiée par: abbigail333

192 km ÷ 64 km/hr = 3 liter

• Réponse publiée par: snow01

60

Step-by-step explanation

G=kd/square root of s

25=100k/sq root of 100

25=100k/10

25×10=100k

250=100k

250/100=k

5/2=k

G=5/2(192)/8

G=480/8

G=60

• Réponse publiée par: axelamat70 liters

Step-by-step explanation:

A variation is a relationship between two variables that share a ratio. This ratio is called "coefficient of variation". It shows how one variable varies as the other.

Some variables are directly related. Direct variation means as one value increase, the other also increases at a rate equal to the coefficient of variation. If 2 variables, x and y are directly related, then their relationship can be stated as k is the coefficient of variation.

Some variables are inversely related. Inverse variation means as one value increase, the other variable decreases at a rate equal to the coefficient of variation. If 2 variables, x and y, are inversely related, their relationship can be stated as There are times when more than 2 variables are related. This is called a joint variation. A joint variation can have direct or indirect variations.

Variations are often solved by getting the value of k.

Let's go back to the problem. Let us assign variables to each quantities.

Let g be the amount of gasoline.

Let d be the distance traveled of the car.

Let s be the speed of the car.

Their relationship can be stated by the variation: We are given that the car used 25 liters for 100 km. on a 100 km/hr speed. Those are values for the variables we have. We can substitute it to the variation and solve for k. Our constant of variation is .

This means we can solve for the variation if we are missing one of variables.

The problem is asking how many liters of gasoline are needed for a 192km trip at a speed of 64 km/hr.

We are given two values (speed and distance), and we have the constant of variation from earlier.  We need liters of gasoline for the trip.

• Réponse publiée par: kimashleybartolome

A.

Step-by-step explanation:

s=d/t

s=200/6

s=33.33 km/h

• Réponse publiée par: snow01

5 s = 10.4 mph

12 s = 60 mph

50 s = 60 mph

83 s = 12.2 mph

Step-by-step explanation:

5 s is between 0 and 12 so use the first model:

5/12×t²

substitute:

5/12×5²=5/12×25=10.4 mph

12 s is between 0 and 12 so use the first model:

5/12×t²

substitute:

5/12×12²=5/12×144=60 mph

50 s is between 12 and 72 so use the 2nd model:

60

therefore, speed is 60 mph

83 s is between 72 and 92 so use tbe 3rd model:

3/20×(92-t)²

substitute:

3/20×(92-83)²=3/20×9²=3/20×81=12.2 mph

• Réponse publiée par: dorothy13
A - amount of gasoline
d = distance travelles
s = speed

A = k(d)(sqrt(s))
25 = k(100)(sqrt(100)) = 1000k
k = 1/40

A = (1/40)(1000)(64)

A = 1600 L
• Réponse publiée par: girly61
EQUATION: A = kd√s

25 = k(100)√(100)

25 = (100k)(10)

25 = 1, 000k

25/1, 000 = 1, 000k/1, 000

1/40 = k <== Constant

A = (1/40)(192)√(64)

A = (24/5)(8)

A = 38.4

• Réponse publiée par: maledabacuetes
EQUATION: A = kd√s

25 = k(100)√(100)

25 = 100k(10)

25/1, 000 = 1, 000k/1, 000

1/40 = k

A = (1/40)(1, 000)√(64)

A = 25(8)

A = 200