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3 years ago

NEED HELP! ASAP! 50 POINTS!

Mathematics

1 answer:

Lerok [7]3 years ago

6 0

Answer:

$y=13\text{ miles per gallon}$

Step-by-step explanation:

We have the two points: (0, 11) and (3, 50).

And we want to find the rate of change in y relative to x, where x is the gallons and y is the miles.

In simple terms, we want to find the slope between the two points.

So, let's use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Let (0, 11) be (x₁, y₁) and let (3, 50) be (x₂, y₂). Substitute. Let's also put "miles" in the numerator with the Ys and "gallons" in the denominator with the Xs. This yields:

$y=\frac{50-11\text{ miles}}{3-0\text{ gallons}}$

Subtract and reduce:

$y=39\text{ miles}/3\text{ gallons}=13\text{ miles per gallon}$

And we're done!

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2 years ago

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Aneli [31]

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Step-by-step explanation: Let's solve your equation step-by-step.

x+ 1 /2

(x−5)=20

Step 1: Simplify both sides of the equation.

x+ 1 /2

(x−5)=20

x+( 1 /2 )(x)+( 1 /2 )(−5)=20(Distribute)

x+ 1 /2

x+ −5 /2 =20

(x+ 1 /2 x)+( −5 /2 )=20(Combine Like Terms)

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Step 2: Add 5/2 to both sides.

3 /2 x + −5 /2 + 5 /2 =20+ 5 /2

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Step 3: Multiply both sides by 2/3.

( 2 /3 )*( 3 /2 x)=( 2 /3 )*( 45 /2 )

x=15

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2 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

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