Solved on Jan 23, 2024

Find the weight (in pounds) of an airplane carrying $45$ gallons of fuel, given that its weight is a linear function of the amount of fuel (in gallons), and the weights are $2028$ pounds with $20$ gallons and $2124$ pounds with $35$ gallons.

STEP 1

Assumptions
1. The weight of the airplane is a linear function of the amount of fuel in its tank.
2. The airplane weighs 2028 pounds with 20 gallons of fuel.
3. The airplane weighs 2124 pounds with 35 gallons of fuel.
4. We are asked to find the weight of the airplane with 45 gallons of fuel.

STEP 2

Let the weight of the airplane be represented by $W$ and the amount of fuel by $F$ . Since the relationship is linear, we can express it as:
$W(F) = mF + b$
where $m$ is the slope of the line and $b$ is the y-intercept, which represents the weight of the airplane with no fuel.

STEP 3

To find the slope $m$ , we use the two given points: $(20, 2028)$ and $(35, 2124)$ . The slope is the change in weight divided by the change in fuel:
$m = \frac{\Delta W}{\Delta F} = \frac{W_2 - W_1}{F_2 - F_1}$

STEP 4

Plug in the given values to calculate the slope $m$ .
$m = \frac{2124 - 2028}{35 - 20}$

STEP 5

Calculate the change in weight and the change in fuel.
$\Delta W = 2124 - 2028 = 96$ $\Delta F = 35 - 20 = 15$

STEP 6

Now calculate the slope $m$ .
$m = \frac{96}{15}$

STEP 7

Simplify the slope $m$ .
$m = \frac{96}{15} = 6.4$

STEP 8

Now we need to find the y-intercept $b$ . We can use one of the points and the slope to solve for $b$ using the equation of the line:
$W = mF + b$

STEP 9

Let's use the point $(20, 2028)$ . Plug in $F = 20$ , $W = 2028$ , and $m = 6.4$ into the equation.
$2028 = 6.4 \cdot 20 + b$

STEP 10

Solve for $b$ .
$2028 = 128 + b$

STEP 11

Subtract 128 from both sides to isolate $b$ .
$b = 2028 - 128$

STEP 12

Calculate the value of $b$ .
$b = 1900$

STEP 13

Now that we have both $m$ and $b$ , we can write the equation of the line representing the weight of the airplane as a function of the amount of fuel.
$W(F) = 6.4F + 1900$

STEP 14

To find the weight of the airplane with 45 gallons of fuel, plug $F = 45$ into the equation.
$W(45) = 6.4 \cdot 45 + 1900$

STEP 15

Calculate the weight with 45 gallons of fuel.
$W(45) = 6.4 \cdot 45 + 1900$

STEP 16

Multiply $6.4$ by $45$ .
$6.4 \cdot 45 = 288$

STEP 17

Add the result to $1900$ .
$W(45) = 288 + 1900$

STEP 18

Calculate the final weight.
$W(45) = 2188$
The airplane weighs 2188 pounds when carrying 45 gallons of fuel.

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Find the weight (in pounds) of an airplane carrying $45$ gallons of fuel, given that its weight is a linear function of the amount of fuel (in gallons), and the weights are $2028$ pounds with $20$ gallons and $2124$ pounds with $35$ gallons.

STEP 1

Assumptions
1. The weight of the airplane is a linear function of the amount of fuel in its tank.
2. The airplane weighs 2028 pounds with 20 gallons of fuel.
3. The airplane weighs 2124 pounds with 35 gallons of fuel.
4. We are asked to find the weight of the airplane with 45 gallons of fuel.

STEP 2

Let the weight of the airplane be represented by $W$ and the amount of fuel by $F$ . Since the relationship is linear, we can express it as:
$W(F) = mF + b$
where $m$ is the slope of the line and $b$ is the y-intercept, which represents the weight of the airplane with no fuel.

STEP 3

To find the slope $m$ , we use the two given points: $(20, 2028)$ and $(35, 2124)$ . The slope is the change in weight divided by the change in fuel:
$m = \frac{\Delta W}{\Delta F} = \frac{W_2 - W_1}{F_2 - F_1}$

STEP 4

Plug in the given values to calculate the slope $m$ .
$m = \frac{2124 - 2028}{35 - 20}$

STEP 5

Calculate the change in weight and the change in fuel.
$\Delta W = 2124 - 2028 = 96$ $\Delta F = 35 - 20 = 15$

STEP 6

Now calculate the slope $m$ .
$m = \frac{96}{15}$

STEP 7

Simplify the slope $m$ .
$m = \frac{96}{15} = 6.4$

STEP 8

Now we need to find the y-intercept $b$ . We can use one of the points and the slope to solve for $b$ using the equation of the line:
$W = mF + b$

STEP 9

Let's use the point $(20, 2028)$ . Plug in $F = 20$ , $W = 2028$ , and $m = 6.4$ into the equation.
$2028 = 6.4 \cdot 20 + b$

STEP 10

Solve for $b$ .
$2028 = 128 + b$

STEP 11

Subtract 128 from both sides to isolate $b$ .
$b = 2028 - 128$

STEP 12

Calculate the value of $b$ .
$b = 1900$

STEP 13

Now that we have both $m$ and $b$ , we can write the equation of the line representing the weight of the airplane as a function of the amount of fuel.
$W(F) = 6.4F + 1900$

STEP 14

To find the weight of the airplane with 45 gallons of fuel, plug $F = 45$ into the equation.
$W(45) = 6.4 \cdot 45 + 1900$

STEP 15

Calculate the weight with 45 gallons of fuel.
$W(45) = 6.4 \cdot 45 + 1900$

STEP 16

Multiply $6.4$ by $45$ .
$6.4 \cdot 45 = 288$

STEP 17

Add the result to $1900$ .
$W(45) = 288 + 1900$

STEP 18

Calculate the final weight.
$W(45) = 2188$
The airplane weighs 2188 pounds when carrying 45 gallons of fuel.

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Find the weight (in pounds) of an airplane carrying 454545 gallons of fuel, given that its weight is a linear function of the amount of fuel (in gallons), and the weights are 202820282028 pounds with 202020 gallons and 212421242124 pounds with 353535 gallons.

STEP 1

Assumptions1. The weight of the airplane is a linear function of the amount of fuel in its tank.2. The airplane weighs 2028 pounds with 20 gallons of fuel.3. The airplane weighs 2124 pounds with 35 gallons of fuel.4. We are asked to find the weight of the airplane with 45 gallons of fuel.

STEP 2

STEP 3

STEP 4

Plug in the given values to calculate the slope m m m.m=2124−202835−20 m = \frac{2124 - 2028}{35 - 20} m=35−202124−2028​

STEP 5

Calculate the change in weight and the change in fuel.ΔW=2124−2028=96 \Delta W = 2124 - 2028 = 96 ΔW=2124−2028=96 ΔF=35−20=15 \Delta F = 35 - 20 = 15 ΔF=35−20=15

STEP 6

Now calculate the slope m m m.m=9615 m = \frac{96}{15} m=1596​

STEP 7

Simplify the slope m m m.m=9615=6.4 m = \frac{96}{15} = 6.4 m=1596​=6.4

STEP 8

Now we need to find the y-intercept b b b. We can use one of the points and the slope to solve for b b b using the equation of the line:W=mF+b W = mF + b W=mF+b

STEP 9

Let's use the point (20,2028) (20, 2028) (20,2028). Plug in F=20 F = 20 F=20, W=2028 W = 2028 W=2028, and m=6.4 m = 6.4 m=6.4 into the equation.2028=6.4⋅20+b 2028 = 6.4 \cdot 20 + b 2028=6.4⋅20+b

STEP 10

Solve for b b b.2028=128+b 2028 = 128 + b 2028=128+b

STEP 11

Subtract 128 from both sides to isolate b b b.b=2028−128 b = 2028 - 128 b=2028−128

STEP 12

Calculate the value of b b b.b=1900 b = 1900 b=1900

STEP 13

Now that we have both m m m and b b b, we can write the equation of the line representing the weight of the airplane as a function of the amount of fuel.W(F)=6.4F+1900 W(F) = 6.4F + 1900 W(F)=6.4F+1900

STEP 14

To find the weight of the airplane with 45 gallons of fuel, plug F=45 F = 45 F=45 into the equation.W(45)=6.4⋅45+1900 W(45) = 6.4 \cdot 45 + 1900 W(45)=6.4⋅45+1900

STEP 15

Calculate the weight with 45 gallons of fuel.W(45)=6.4⋅45+1900 W(45) = 6.4 \cdot 45 + 1900 W(45)=6.4⋅45+1900

STEP 16

Multiply 6.4 6.4 6.4 by 45 45 45.6.4⋅45=288 6.4 \cdot 45 = 288 6.4⋅45=288

STEP 17

Add the result to 1900 1900 1900.W(45)=288+1900 W(45) = 288 + 1900 W(45)=288+1900

STEP 18

Calculate the final weight.W(45)=2188 W(45) = 2188 W(45)=2188The airplane weighs 2188 pounds when carrying 45 gallons of fuel.

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Find the weight (in pounds) of an airplane carrying $45$ gallons of fuel, given that its weight is a linear function of the amount of fuel (in gallons), and the weights are $2028$ pounds with $20$ gallons and $2124$ pounds with $35$ gallons.

Assumptions
1. The weight of the airplane is a linear function of the amount of fuel in its tank.
2. The airplane weighs 2028 pounds with 20 gallons of fuel.
3. The airplane weighs 2124 pounds with 35 gallons of fuel.
4. We are asked to find the weight of the airplane with 45 gallons of fuel.

Plug in the given values to calculate the slope $m$ .
$m = \frac{2124 - 2028}{35 - 20}$

Calculate the change in weight and the change in fuel.
$\Delta W = 2124 - 2028 = 96$ $\Delta F = 35 - 20 = 15$

Now calculate the slope $m$ .
$m = \frac{96}{15}$

Simplify the slope $m$ .
$m = \frac{96}{15} = 6.4$

Now we need to find the y-intercept $b$ . We can use one of the points and the slope to solve for $b$ using the equation of the line:
$W = mF + b$

Let's use the point $(20, 2028)$ . Plug in $F = 20$ , $W = 2028$ , and $m = 6.4$ into the equation.
$2028 = 6.4 \cdot 20 + b$

Solve for $b$ .
$2028 = 128 + b$

Subtract 128 from both sides to isolate $b$ .
$b = 2028 - 128$

Calculate the value of $b$ .
$b = 1900$

Now that we have both $m$ and $b$ , we can write the equation of the line representing the weight of the airplane as a function of the amount of fuel.
$W(F) = 6.4F + 1900$

To find the weight of the airplane with 45 gallons of fuel, plug $F = 45$ into the equation.
$W(45) = 6.4 \cdot 45 + 1900$

Calculate the weight with 45 gallons of fuel.
$W(45) = 6.4 \cdot 45 + 1900$

Multiply $6.4$ by $45$ .
$6.4 \cdot 45 = 288$

Add the result to $1900$ .
$W(45) = 288 + 1900$

Calculate the final weight.
$W(45) = 2188$
The airplane weighs 2188 pounds when carrying 45 gallons of fuel.