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n200080 [17]
3 years ago
8

A slide 4.1 meters long makes an angle of 35º with the ground. To the nearest tenth of a meter, how far above the ground is the

top of the slide?
Mathematics
1 answer:
Step2247 [10]3 years ago
4 0

length of the slide is lets say 'l' the height (h) opposite to the angle 28 is found by h=l*sin28 where sin28=0.47 approximately h=4.5*0.47 h=2.115m

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Answer:

Step-by-step explanation:

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Combine like terms. 1/4 and 7/8 are like terms and LCD of 4 and 8 is 8

9/10y and -3/5y are like terms and LCD of 10 , 5 is 10

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Step-by-step explanation:

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Find the rate of change of the function h(x) = 2 x on the interval 2 ≤ x ≤ 4
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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

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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)

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

∵ h(0) = 144 and v = 128 ft/sec

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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

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

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c. Attached graph

19)

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∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

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

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