The target temperature for a laboratory is 72° Fahrenheit. It
1 answer:
The inequality |t - 72| ≤ 2 represents the target temperature for a laboratory is 72° Fahrenheit, and it can vary by up to 2° option (D) is correct.
<h3>What is inequality?</h3>It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than, known as inequality.
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We have:
The target temperature for a laboratory is 72° Fahrenheit. It can vary by up to 2°.
Target temperature = 72 degree Fahrenheit
It can vary by up to 2°. (the highest limit)
The temperature in the laboratory represented by t
Now,
inequality represents this situation is:
|t - 72| ≤ 2
Thus, the inequality |t - 72| ≤ 2 represents the target temperature for a laboratory is 72° Fahrenheit, and it can vary by up to 2° option (D) is correct.
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<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer