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3 years ago

15

NO LINKS OR ASSESSMENT!! Part 2a: State the Domain and Range of each graph in BOTH inequality and interval notation

1 answer:

ad-work [718]3 years ago

7 0

9514 1404 393

Answer:

see below

Step-by-step explanation:

The domain is the horizontal extent; the range is the vertical extent. Open circles indicate points not included in either. Any point on the curve is included in the domain and range.

__

1. D: -2 < x ≤ 2, (-2, 2]

R: -4 < y ≤ 4, (-4, 4]

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2. D: -5 ≤ x ≤ 1, [-5, 1]

R: -1 ≤ y ≤ 5, [-1, 5]

__

3. D: -4 < x < 4, (-4, 4)

R: -1 ≤ y < 3, [-1, 3)

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3 years ago

NGC is a hot air balloon in the sky from her spot on the ground. The angle of elevation from Angie to the balloon is 40°. If she

Goshia [24]

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The distance from hot air balloon to the ground $AC=50.1457 ft$ .

Step-by-step explanation:

Labelled diagram of given scenario is shown below.

Given that,

An angle of elevation of Hot air balloon by Angie is $40$ °.

When she stepped back $200 ft$ then angle of elevation was $10$ °.

Height of Angie is $5.5 ft$ .

To find: How far off the ground is a hot air balloon.

So, from figure

Height of Angie $(BC) = 5.5ft$

In triangle Δ $ABD$ ,

⇒ $tan {40\si {\degree}} = \frac{AB}{BD}$

⇒ $AB = tan (40) \times BD$ ....................(1)

Now, In triangle Δ $ABE$

⇒ $tan (10)= \frac{AB}{200+BD}$

⇒ $AB = tan(10)\times (200+BD)$

Here, substituting the value $AB$ from Equation (1) we get,

$tan(10)\times (200+BD)= tan(40)\times BD$

$tan(10)\times 200+tan(10)\times BD= tan(40)\times BD$

⇒ $BD\times (tan(40)-tan(10))=tan(10)\times 20$

⇒ $BD\times 0.6628 = 35.2654$

⇒ $BD = \frac{35.2654}{0.6628} =53.2067 ft$

Now, finding the value of $AB$ from equation (1)

$AB= tan(40)\times 53.2067=0.8391\times 53.2067$

$AB=44.6457 ft$

Therefore Length of $AC= AB+BC$ = $(44.6457 + 5.5) ft$

$AC=50.1457 ft$ .

Hence,

The distance from hot air balloon to the ground $AC=50.1457 ft$ .

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