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Gemiola [76]
3 years ago
13

An observer on the ground is x meters from the base of the launch pad of a rocket, which is at the same level as the observer. A

few seconds after the rocket takes off vertically, the observer sees its tip at an angle of q° from the horizontal. How far above the ground is the tip of the rocket at that instant? Assume that the ground is level.
x tan q
x sin q
x cos q

Mathematics
2 answers:
kondaur [170]3 years ago
7 0
Hello;

Answer A

h/x= tan q=>h=tan q* x

weeeeeb [17]3 years ago
7 0

Answer:

Tip of the rocket is x tan q meters above the ground.

Step-by-step explanation:

From the figure attached, at any moment after the launch of the rocket, tip of the rocket is at A.

AB is the height of the rocket tip from the ground and observer is x meters away from the launch pad.

Then tan q = \frac{AB}{BC}

         tan q = \frac{AB}{x}

AB = x tan q meters.

Tip of the rocket is x tan q meters above the ground.

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Answer:

How much tax would you pay on a new TV that is $899 at Best Buy, if the tax rate is 6.5%?. answer choices. $957.44. $840.00. $58.44.

Step-by-step explanation:

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3 years ago
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Which is traveling faster, a car whose velocity vector is 201 + 25), or a car whose velocity vector is 30i, assuming that the un
Snowcat [4.5K]

Answer:

2nd car is running faster than the first car by 2.01 units.

Step-by-step explanation:

Let's assume that

  • the velocity of first car,

v_1\ =\ 20i\ +\ 25j

  • and the velocity of second car

v_2\ =\ 30i

=> speed of first car,

u_1\ =\ \sqrt{(20)^2+(25)^2}

      =\ \sqrt{400+625}

       =\ \sqrt{1025}

       = 32.01 units

and speed of second car,

u_2\ =\ \sqrt{(30)^2+(0)^2}

        = 30 units

u_2\ -\ u_1\ =\ 32.01\ -30

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Hence, 2nd car is running faster than the first car by 2.01 units.

8 0
3 years ago
What is the sum?<br> 2/2x^+4/x^2
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Answer:

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Step-by-step explanation:

6 0
3 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
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So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
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Step-by-step explanation:

uh no explanation needed

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