Soru

Zorluk: OrtaBasic Algebraic Expressions and One-Step Equations

During a laboratory experiment, a liquid cooling solution's temperature dropped by a total of 14.4C14.4^\circ\text{C} over a duration of tt minutes. If the temperature decreased at a constant rate of 1.8C1.8^\circ\text{C} per minute, which of the following equations can be solved to find tt?

  1. 1.8t=14.41.8t = 14.4Cevap
  2. B
    t1.8=14.4\frac{t}{1.8} = 14.4
  3. C
    t1.8=14.4t - 1.8 = 14.4
  4. D
    t+1.8=14.4t + 1.8 = 14.4
  5. E
    14.4t=1.814.4t = 1.8

Cevap

The equation 1.8t=14.41.8t = 14.4 correctly models the situation.
Because the cooling rate is constant at 1.8C1.8^\circ\text{C} per minute, the total temperature change over tt minutes is found by multiplying the rate by time (1.8t1.8 \cdot t). Setting this equal to the total drop of 14.4C14.4^\circ\text{C} gives 1.8t=14.41.8t = 14.4.

Adım Adım Çözüm

1
Identify the given quantities and their relationship.
Rate of change = 1.8C/min1.8^\circ\text{C/min}, Time = tt minutes, Total change = 14.4C14.4^\circ\text{C}.
The total change in a constant rate scenario is the product of the rate and the time duration.
2
Set up the one-step algebraic equation.
Rate×Time=Total Change    1.8×t=14.4\text{Rate} \times \text{Time} = \text{Total Change} \implies 1.8 \times t = 14.4
Translating the verbal relationship into an algebraic equation yields 1.8t=14.41.8t = 14.4.

Anahtar Kavram

Translating constant rate scenarios into one-step linear multiplication equations
Tahmini Süre:1m 0s
Bu soruyu puanla