34 points nita finds that the expression (a+1)^2 is sometimes, but not always, greater in value than (a+1)^3 when she evaluates both expressions for the same value of a. how do you explain this finding?

Consider the two expressions, (a + 1)² and (a + 1)³.

No matter what the number is, a² ≥ 0. Even if you square a negative number, you still get a positive number. We include the "or equal to" since 0² = 0. Since a² ≥ 0 is always true, then we can manipulate this inequality.

a² ≥ 0

a² + 2a + 1 ≥ 0 + 2a +1 by adding 2a +1 to both sides

(a + 1)² ≥ 2a + 1 as a² + 2a + 1 factors into (a +1)²

(a + 1)² ≥ 1 If it's more than 2a +1, it has to be more than 1

(a + 1)² ≥ 0 If it's more than one, it's more than zero since 1 > 0.

So we can conclude that (a + 1) ≥ 0 is always true.

Now let's look at (a + 1)³. We know that from before (a + 1)² ≥ 0. A tempting thing to say is that if you multiply by both sides by (a + 1) then both will be more than zero. Doing so isn't right.

What we instead must do is use cases. Either a is positive, negative, or zero.

Case 1: a = 0

When a = 0, (a + 1)³ = 1³ = 1 and 1 > 0. Thus (a + 1)³ ≥ 0,

Case 2: a > 0

When a > 0, a² > 0 from multiplying both sides by a. Do it a second time and a³ > 0. Then, if we add terms to both sides like in the a² example, we have this:

a³ > 0

a³ + 3a² > 3a² by adding 3a² to both sides

a³ + 3a² + 3a > 3a² + 3a by adding 3a to both sides

a³ + 3a² + 3a + 1 > 3a² + 3a + 1 by adding 1 to both sides

(a + 1)³ > 3a² + 3a + 1 by factoring the left side

(a + 1)³ > 1 Since a > 0 by assumption then 3a > 0 and 3a² > 0 and their sum is more than zero too

(a + 1)³ > 0 Since 1 > 0

Case 3: a < 0

Since a < 0 then a² > 0 (minus times minus is plus), but a³ < 0 through a similar multiplication.

Let b = a + 1. If a < 0 then a + 1 < 1 by adding 1 to both sides and b < 1 by back substitution. So b³ < 1³ by cubing both sides and b³ < 1 since 1³ =1.

b³ < 1

(a+1)³ < 1 We chose a + 1 = b.

When a < 0, we can conclude that (a + 1)³ < 1. When a ≥ 0, then (a +1)³ ≥ 0.

However for all a, (a +1)² ≥ 0. Thus, we have our sometimes truth. That when we choose a to be negative we have that the (a + 1)² and (a + 1)³ are of opposite signs.

Consider the two expressions, (a + 1)² and (a + 1)³.No matter what the number is, a² ≥ 0. Even if you square a negative number, you still get a positive number. We include the "or equal to" since 0² = 0. Since a² ≥ 0 is always true, then we can manipulate this inequality.a² ≥ 0a² + 2a + 1 ≥ 0 + 2a +1 by adding 2a +1 to both sides(a + 1)² ≥ 2a + 1 as a² + 2a + 1 factors into (a +1)²(a + 1)² ≥ 1 If it's more than 2a +1, it has to be more than 1(a + 1)² ≥ 0 If it's more than one, it's more than zero since 1 > 0.So we can conclude that (a + 1) ≥ 0 is always true.Now let's look at (a + 1)³. We know that from before (a + 1)² ≥ 0. A tempting thing to say is that if you multiply by both sides by (a + 1) then both will be more than zero. Doing so isn't right.What we instead must do is use cases. Either a is positive, negative, or zero.Case 1: a = 0When a = 0, (a + 1)³ = 1³ = 1 and 1 > 0. Thus (a + 1)³ ≥ 0,Case 2: a > 0When a > 0, a² > 0 from multiplying both sides by a. Do it a second time and a³ > 0. Then, if we add terms to both sides like in the a² example, we have this:a³ > 0a³ + 3a² > 3a² by adding 3a² to both sidesa³ + 3a² + 3a > 3a² + 3a by adding 3a to both sidesa³ + 3a² + 3a + 1 > 3a² + 3a + 1 by adding 1 to both sides(a + 1)³ > 3a² + 3a + 1 by factoring the left side(a + 1)³ > 1 Since a > 0 by assumption then 3a > 0 and 3a² > 0 and their sum is more than zero too(a + 1)³ > 0 Since 1 > 0Case 3: a < 0Since a < 0 then a² > 0 (minus times minus is plus), but a³ < 0 through a similar multiplication.Let b = a + 1. If a < 0 then a + 1 < 1 by adding 1 to both sides and b < 1 by back substitution. So b³ < 1³ by cubing both sides and b³ < 1 since 1³ =1.b³ < 1(a+1)³ < 1 We chose a + 1 = b.When a < 0, we can conclude that (a + 1)³ < 1. When a ≥ 0, then (a +1)³ ≥ 0.However for all a, (a +1)² ≥ 0. Thus, we have our sometimes truth. That when we choose a to be negative we have that the (a + 1)² and (a + 1)³ are of opposite signs.

Step-by-step explanation:

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."