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A charging station can only charge cell phones and laptops. Cell phones use approximately 3 joules per second ((J)/(s)) when charging, and laptops use approximately 40(J)/(s) when charging. Which of the following equations represents the relationship between the number of cell phones, c, and the number of laptops, l, that could be charging when the station's output is exactly 160(J)/(s) ? Choose 1 answer: (A) c=-40 l+160 (B) c=-(40)/(3)l+(160)/(3) (c) c=-3l+160 (D) c=-(3)/(40)l+(160)/(40)

A charging station can only charge cell phones and laptops. Cell phones use approximately 33 joules per second (Js) \left(\frac{\mathrm{J}}{\mathrm{s}}\right) when charging, and laptops use approximately 40Js 40 \frac{\mathrm{J}}{\mathrm{s}} when charging. Which of the following equations represents the relationship between the number of cell phones, c c , and the number of laptops, l l , that could be charging when the station's output is exactly 160Js 160 \frac{\mathrm{J}}{\mathrm{s}} ?\newlineChoose 11 answer:\newline(A) c=−40l+160 c=-40 l+160 \newline(B) c=−403l+1603 c=-\frac{40}{3} l+\frac{160}{3} \newline(C) c=−3l+160 c=-3 l+160 \newline(D) c=−340l+16040 c=-\frac{3}{40} l+\frac{160}{40}

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Q. A charging station can only charge cell phones and laptops. Cell phones use approximately 33 joules per second (Js) \left(\frac{\mathrm{J}}{\mathrm{s}}\right) when charging, and laptops use approximately 40Js 40 \frac{\mathrm{J}}{\mathrm{s}} when charging. Which of the following equations represents the relationship between the number of cell phones, c c , and the number of laptops, l l , that could be charging when the station's output is exactly 160Js 160 \frac{\mathrm{J}}{\mathrm{s}} ?\newlineChoose 11 answer:\newline(A) c=−40l+160 c=-40 l+160 \newline(B) c=−403l+1603 c=-\frac{40}{3} l+\frac{160}{3} \newline(C) c=−3l+160 c=-3 l+160 \newline(D) c=−340l+16040 c=-\frac{3}{40} l+\frac{160}{40}
  1. Define Total Power: Let's denote the total power used by cell phones as PcP_c and the total power used by laptops as PlP_l. The power used by cell phones is 33 joules per second per cell phone, and the power used by laptops is 4040 joules per second per laptop. The total power output of the station is 160160 joules per second. We can write the equation representing the total power output as:\newlinePc+Pl=160P_c + P_l = 160\newlineSubstituting the power usage for cell phones and laptops, we get:\newline3c+40l=1603c + 40l = 160
  2. Equation Representation: We need to express cc in terms of ll. To do this, we will isolate cc on one side of the equation. We can subtract 40l40l from both sides to get:\newline3c=160−40l3c = 160 - 40l
  3. Isolate cc: Now, we divide both sides of the equation by 33 to solve for cc:c=160−40l3c = \frac{160 - 40l}{3}
  4. Solve for c: We can simplify the equation by distributing the division across the terms in the numerator: c=1603−40l3c = \frac{160}{3} - \frac{40l}{3}
  5. Final Simplification: The equation c=1603−40l3c = \frac{160}{3} - \frac{40l}{3} matches one of the given choices, which is:\newline(B) \(c = -\left(\frac{\(40\)}{\(3\)}\right)l + \left(\frac{\(160\)}{\(3\)}\right)

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