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icang [17]
3 years ago
15

A time/motion word problem

Mathematics
2 answers:
andriy [413]3 years ago
3 0
The two planes are flying on the two legs of a right triangle.
The straight distance between them is the hypotenuse of the triangle.

Since the speeds are in mph, let's work the time in hours.
Call the time 'H' that we're looking for.
It's the number of hours after they both take off that they're 650 miles apart.

After 'H' hours, the first plane has gone 500H miles north.
After 'H' hours, the second plane has gone 1200H miles east.
After 'H' hours, they are 650 miles apart.

Do you remember this for a right triangle ? ==>    A² + B² = C²

(500H)² + (1200H)² = (650)²

250,000H² + 1,440,000H² = 422,500

1,690,000 H² = 422,500

H² = (422,500) / (1,690,000) = 0.25

H = √0.25 = 1/2 hour = 30 minutes
bezimeni [28]3 years ago
3 0
x^2=500^2+1200^2\\\\x^2=250000+1440000\ \ \ \Rightarrow\ \ \ \ x^2=1690000\ \ \ \Rightarrow\ \ \ \ x=1300\ [mph]\\\\the\ distance=650\ miles\\\\the\ speed= \frac{the\ distance}{the\ time} \ \ \ \Rightarrow\ \ \ 1300\ [mph]= \frac{650\ [miles]}{the\ time}\\\\the\ time= \frac{650}{1300} \ [hr]=0.5\ [hr]=30\ [min]\\\\Ans.\ after\ 30\ minutes.

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solmaris [256]

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7 0
2 years ago
Find four consecutive integers with a sum of 106.
grandymaker [24]
4 consecutive integers...
x, x + 1, x + 2, x + 3

x + x + 1 + x + 2 + x + 3 = 106
4x + 6 = 106
4x = 106 - 6
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6 0
3 years ago
A business executive bought 40 stamps for
alekssr [168]

Answer:

32 33-cent stamps, 8 23-cent stamps

Step-by-step explanation:

In order to solve this question, we need to set up a system of equations. Also known as solving for two variables (the number of each stamp).

Let's set x to be the number of 33-cent stamps. Similarly, let's set y to be the number of 23-cent stamps.

To make our first equation, let's think about the number of stamps total we have. We can say:

x + y =40

<em>(AKA - The number of 33-cent stamps, plus the number of 23-cent stamps, equals 40 stamps.)</em>

Now, let's make an equation for the cost of these stamps.

0.33x + 0.23y = 12.40

<em>(AKA - The cost of the stamps in total, should equal $12.40).</em>

So now, we have our two equations:

x + y =40\\0.33x+0.23y=12.40

If you have a TI-84 graphing calculator, you can go to apps -> polysmlt2 -> simultaneous eqn solver, and then input these equations into the menu. This will solve the problem for you.

If you need to do this manually, let's use substitution. Condense our first equation to make it more substitutable.

x+y=40\\x=40-y

Now, let's put this into our second equation.

0.33x+0.23y=12.40\\0.33(40-y)+0.23y=12.40

Distribute, and solve for y.

13.2-0.33y +0.23y =12.40\\13.2-0.1y=12.40\\-0.1y=-0.8\\y=8

Now, we plug this into one of our equations.

x+y=40\\x+8=40\\x=32

In the end, we have thirty-two 33-cent stamps, and eight 23-cent stamps.

3 0
1 year ago
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