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kifflom [539]
3 years ago
7

Two trains leave the same station at the same time, but are headed in opposite directions. One train travels at 70 mph and the o

ther train travels at 80 mph. How much times passes until the trains are 600 mi apart?
Mathematics
2 answers:
lidiya [134]3 years ago
7 0
Imagine you're on the first train. since you're moving away from second train at 70mph, and that train is moving away from you at 80mph, in total that other train for you is moving away at 150mph (as you are stationary inside the train).

if the two trains are moving apart at 150mph, then to be 600 miles apart it will take 600/150 hours, which you can calculate :)
Genrish500 [490]3 years ago
3 0

Answer:

Time = 4 hours.

Step-by-step explanation:

Since, both the train travels with the speed 70 and 80 mph from the same points in opposite direction.

Thus, we can take speed of train = 70 + 80 = 150 mph

Also, after some time distance between both train is 600 m

Thus, using the formula,

\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}

⇒ \text{Time}=\dfrac{\text{Distance}}{\text{Speed}}

⇒ \text{Time}=\dfrac{\text{600}}{\text{150}}

⇒ Time = 4 hours.

Thus, After 4 hours both train will 600 m apart.

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5 0
3 years ago
Patrick has just finished building a pen for his new dog. The pen is 3 feet wider than it is long. He also built a doghouse to p
zalisa [80]

General Idea:

(i) Assign variable for the unknown that we need to find

(ii) Sketch a diagram to help us visualize the problem

(iii) Write the mathematical equation representing the description given.

(iv) Solve the equation by substitution method. Solving means finding the values of the variables which will make both the equation TRUE

Applying the concept:

Given: x represents the length of the pen and y represents the area of the doghouse

<u>Statement 1: </u>"The pen is 3 feet wider than it is long"

Length \; of\; the \; pen = x\\ Width \; of\; the\; pen=x+3

------

<u>Statement 2: "He also built a doghouse to put in the pen which has a perimeter that is equal to the area of its base"</u>

Area \; of\; the\; Dog \; house=y\\ Perimeter \; of\; Dog\; house=y

------

<u>Statement 3: "After putting the doghouse in the pen, he calculates that the dog will have 178 square feet of space to run around inside the pen."</u>

Area \; of \; the\; Pen - Area \;of \;the\;Dog \;House=\;Space\;inside\;Pen\\ \\ x \cdot (x+3)-y=178\\ Distributing \;x\;in\;the\;left\;side\;of\;the\;equation\\ \\ x^2+3x-y=178\Rightarrow\; 1^{st}\; Equation\\

------

<u>Statement 4: "The perimeter of the pen is 3 times greater than the perimeter of the doghouse."</u>

Perimeter\; of\; the\; Pen=3\; \cdot \; Perimeter\; of\; the\; Dog\; House\\ \\ 2(x \; + \; x+3)=3 \cdot y\\ Combine\; like\; terms\; inside\; the\; parenthesis\\ \\ 2(2x+3)=3y\\ Distribute\; 2\; in\; the\; left\; side\; of\; the\; equation\\ \\ 4x+6=3y\\ Subtract \; 6\; and \; 3y\; on\; both\; sides\; of\; the\; equation\\ \\ 4x+6-3y-6=3y-3y-6\\ Combine\; like\; terms\\ \\ 4x-3y=-6 \Rightarrow \; \; 2^{nd}\; Equation\\

Conclusion:

The systems of equations that can be used to determine the length and width of the pen and the area of the doghouse is given in Option B.

178=x^2+3x-y\\ \\ -6=4x-3y

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