5) The less than symbol, using x, is <em>x < __</em>. The blank should have the number that comes after, which is <em>-15°F</em>. So, that will look like x < -15°F
6) When something is at most __, that means that __ is the greatest number, which means the inequality looks like <em>x ≤ __</em>. The number that goes into the blank is <em>127 pounds</em>, which makes the whole inequality x ≤ 127 pounds.
7) An inequality is solved similarly to an equation, with a few exceptions.
x - 3 < -15
Subtract: x < -12
That looks like what is shown in the image. The circle is empty because the inequality does not say that it contains -12, so it is skipped. It goes to the left because the inequality shows all numbers less than -12.
Answer:
1: (-24, -2, 6) 2: June 3:January
Step-by-step explanation:
1: the closer the negatives get to 0 backwards, the higher the temperature is, and after 0 would be even higher.
2: -24 is the most far back from the other two
3: 2 of them are negatives and the other is positive so the positive number would be the highest temp
Answer:
-8 and 6
Step-by-step explanation:
Coordinate of point M = -1
Distance between M and a point N = MN = 7
Therefore, the possible coordinates of N can either be:
To the left => -1 - 7 = -8
Or
To the right => -1 + 7 = 6
Check:
If M = -1, and N = -8, MN = |-1 -(-8)| = |-1 + 8| = 7
Or
If M = -1, and N = 6, MN = |-1 - 6| = |-7| = 7
So, our answer is right.
Possible coordinates of point N are -8 and 6
This is so provided that the velocity changes continuously in which case we can apply the mean value theorem.
<span>Velocity (v) is the derivative of displacement (x) : </span>
<span>v = dx/dt </span>
<span>Monk 1 arrives after a time t* and Monk 2 too. </span>
<span>Name v1(t) and v2(t) their respective velocities throughout the trajectory. </span>
<span>Then we know that both average velocities were equal : </span>
<span>avg1 = avg2 </span>
<span>and avg = integral ( v(t) , t:0->t*) / t* </span>
<span>so </span>
<span>integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*) </span>
<span>which is the same of saying that the covered distances after t* seconds are the same </span>
<span>=> integral (v1(t) - v2(t) , t:0->t*) = 0 </span>
<span>Thus, name v#(t) = v1(t) - v2(t) , then we obtain </span>
<span>=> integral ( v#(t) , t:0->t*) = 0 </span>
<span>Name the analytical integral of v#(t) = V(t) , then we have </span>
<span>=> V(t*) - V(0) = 0 </span>
<span>=> V(t*) = V(0) </span>
<span>So there exist a c in [0, t*] so that </span>
<span>V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem) </span>
<span>We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get </span>
<span>V'(c) = v#(c) = v1(c) - v2(c) = 0 </span>
<span>=> v1(c) = v2(c) </span>
<span>So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2. </span>