Question

the temperature yesterday in perrysburg, ohio, was $-6^circ$f. the temperature today in perrysburg is $6^circ$f warmer. what is the temperature today? a $-12^circ$f b $12^circ$f c $6^circ$f d $0^circ$f

Explanation:

Step1: Identify yesterday's temp

Yesterday's temp: $-6^\circ\text{F}$

Step2: Add today's increase

Today's temp = $-6 + 6$

Step3: Calculate result

$-6 + 6 = 0^\circ\text{F}$? Wait, no—wait, wait, no: Wait, the problem says "6°F warmer". So starting from -6, adding 6. So $-6 + 6 = 0$? Wait, no, wait—wait, maybe I miscalculated. Wait, -6 + 6 is 0? Wait, but let's check again. Wait, the options have D as 0°F. Wait, let's re-express:

Yesterday: $-6^\circ\text{F}$

Today: warmer by $6^\circ\text{F}$, so we add 6 to yesterday's temp.

So calculation: $-6 + 6 = 0^\circ\text{F}$. Wait, but let's confirm. So if it was -6, and it's 6 degrees warmer, we move 6 units to the right on the number line. From -6, moving 6 right: -6 +1= -5, +2=-4, +3=-3, +4=-2, +5=-1, +6=0. So yes, 0°F. Wait, but the options: D is 0°F. Wait, but let me check the options again. The options are A: -12, B:12, C:6, D:0. So the correct calculation is -6 +6 =0. So the answer is D.

Wait, but maybe I made a mistake? Wait, no. Let's re-express:

Temperature today = yesterday's temperature + increase in temperature.

Yesterday's temperature: $T_{y} = -6^\circ\text{F}$

Increase: $\Delta T = 6^\circ\text{F}$

So $T_{today} = T_y + \Delta T = -6 + 6 = 0^\circ\text{F}$

Yes, that's correct. So the answer is D.

Answer:

D. $0^\circ\text{F}$