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Stolb23 [73]
3 years ago
14

PLEASE HELP 34 POINTS

Mathematics
2 answers:
Taya2010 [7]3 years ago
7 0

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.

maks197457 [2]3 years ago
3 0

Answer:

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:

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Step-by-step explanation:

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1. A plane is flying at an altitude of 13000 feet. The angle of depression from the plane to a control tower is 20 degrees. The
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Answer:

see explanation

Step-by-step explanation:

There is a right triangle formed between the height above the tower to the plane, the horizontal and the line from the tower to the plane.

The height above the tower to the plane = 13000 - 150 = 12850 ft

The angle between the plane and the tower is 20° , angle of depression.

Using the tangent ratio in the right triangle formed, with the adjacent side being the horizontal ( hor ) distance, then

tan20° = \frac{opposite}{adjacent} = \frac{12850}{hor} ( multiply both sides by hor)

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Arianna is 60 + 5 = 65 ft above the ground ( to her eye level )

Using the right triangle formed with legs 65 and 37  and angle of depression x, then

tanx° = \frac{opposite}{adjacent} = \frac{37}{65} , thus

x = tan^{-1} (\frac{37}{65} ) ≈ 30° ( to the nearest degree )

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Read 2 more answers
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