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D=27.4 m+12,100
The distance, 
D, in millions of miles, of the Voyager 1 spacecraft from the Sun 
m months after January 1,2015 is approximated by the equation. What is the best interpretation of 27.4 as shown in the given equation?
Choose 1 answer:
(A) Voyager 1 left Earth 27.4 months before January 1 , 2015.
(B) The Voyager 1 has a speed of 27.4 million miles per hour.
(C) The Voyager 1 was a distance of 27.4 million miles from the Sun on January 1, 2015.
(D) The distance between Voyager 1 and the Sun increases by 27.4 million miles every month.
linear relationships | Lesson

D=27.4m+12,100 D=27.4 m+12,100 \newlineThe distance, D D , in millions of miles, of the Voyager 11 spacecraft from the Sun m m months after January 11,20152015 is approximated by the equation. What is the best interpretation of 2727.44 as shown in the given equation?\newlineChoose 11 answer:\newline(A) Voyager 11 left Earth 2727.44 months before January 11 , 20152015.\newline(B) The Voyager 11 has a speed of 2727.44 million miles per hour.\newline(C) The Voyager 11 was a distance of 2727.44 million miles from the Sun on January 11, 20152015.\newline(D) The distance between Voyager 11 and the Sun increases by 2727.44 million miles every month.\newlinelinear relationships | Lesson

Full solution

Q. D=27.4m+12,100 D=27.4 m+12,100 \newlineThe distance, D D , in millions of miles, of the Voyager 11 spacecraft from the Sun m m months after January 11,20152015 is approximated by the equation. What is the best interpretation of 2727.44 as shown in the given equation?\newlineChoose 11 answer:\newline(A) Voyager 11 left Earth 2727.44 months before January 11 , 20152015.\newline(B) The Voyager 11 has a speed of 2727.44 million miles per hour.\newline(C) The Voyager 11 was a distance of 2727.44 million miles from the Sun on January 11, 20152015.\newline(D) The distance between Voyager 11 and the Sun increases by 2727.44 million miles every month.\newlinelinear relationships | Lesson
  1. Understand Coefficient Role: We are given the equation D=27.4m+12,100D=27.4m+12,100, where DD is the distance in millions of miles and mm is the number of months after January 11, 20152015. To interpret the coefficient 27.427.4, we need to understand its role in the equation.
  2. Linear Equation Interpretation: The equation is linear, which means that for every increase in mm, the distance DD increases by the coefficient of mm. In this case, the coefficient is 27.427.4, which means that for every month, the distance increases by 27.427.4 million miles.
  3. Option Evaluation A: Now, let's evaluate the options given:\newline(A) This option suggests that Voyager 11 left Earth 27.427.4 months before January 11, 20152015. This interpretation is not supported by the equation, as the coefficient 27.427.4 is multiplied by the number of months after January 11, 20152015, not before.
  4. Option Evaluation B: (B) This option suggests that Voyager 11 has a speed of 27.427.4 million miles per hour. However, the equation does not mention hours, and the coefficient is associated with the number of months, not hours. Therefore, this interpretation is incorrect.
  5. Option Evaluation C: (C) This option suggests that Voyager 11 was a distance of 27.427.4 million miles from the Sun on January 11, 20152015. This interpretation is incorrect because the constant term in the equation (12,10012,100) represents the initial distance from the Sun on January 11, 20152015, not the coefficient 27.427.4.
  6. Option Evaluation D: (D) This option suggests that the distance between Voyager 11 and the Sun increases by 27.427.4 million miles every month. This interpretation is consistent with the role of the coefficient in the linear equation, which indicates the rate of change of the distance with respect to time (months).

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