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ioda
2 years ago
9

To input your angle measures in degrees, use the functions sind, cosd and tand, instead of sin, cos and tan. Zach stands at the

top of a cliff. On the ground, 310 feet below, Zach spots a coyote. When Zach initially sees the coyote, the angle of depression for his vision is 6 degrees. Draw a picture! At this point in time, how far is the coyote from the base of the cliff
Mathematics
1 answer:
kramer2 years ago
6 0

Answer:

  about 2949 feet

Step-by-step explanation:

The geometry of the situation can be modeled by a right triangle. The height of the cliff can be taken to be the side opposite the given angle, and the distance to the coyote will be the side adjacent to the given angle. The relation between these values is the trig function ...

  Tan = Opposite/Adjacent

__

<h3>setup</h3>

Filling in the known values, we have ...

  tan(6°) = (310 ft)/(distance to coyote)

<h3>solution</h3>

Multiplying by (distance to coyote)/tan(6°) gives ...

  distance to coyote = (310 ft)/tan(6°) ≈ 310/0.105104 ft

  distance to coyote ≈ 2949.453 ft

The coyote is about 2949 feet from the base of the cliff.

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Find the common ratio of the geometrie sequence 6, 24, 96,
inessss [21]

Answer:

4

Step-by-step explanation:

To find the common ratio, take the second term and divide by the first term

24/6 = 4

Check by taking the third term and dividing by the second term

96/24 = 4

The common ratio is 4

4 0
2 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

8 0
2 years ago
Help me pleaseeeeeeeeeeeeee
lubasha [3.4K]
I had this and it was b for me
6 0
2 years ago
4,3,5,9,12,17,...what is the next number?
77julia77 [94]

Answer:

The next number is going to be 21

6 0
2 years ago
Read 2 more answers
Como se hace el calculo de: 12 + 9a + 18b = ----- * ( ---- + 3a + ----- )
nexus9112 [7]

Answer:

3(4+3a+6b)

Step-by-step explanation:

8 0
3 years ago
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