the angle of elevation of a jet plane from a point P on the ground is 60 degree after a flight of 15 seconds the angle of elevat
ion changes to 30 degree if the jet plane is flying at a constant height of 1500 root 3 metre find the speed of the jet plane
1 answer:
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For this case, the parent function is given by:
We apply the following function transformation:
Vertical compressions:
To graph y = a * f (x)
If 0 <a <1, the graph of y = f (x) is compressed vertically by a factor a. (Shrinks)
We have then:
Horizontal translations:
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
Vertical translations:
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then:
Answer:
Vertical compression by factor of 1/2. Horizontal displacement 3 units to the left. Vertical displacement 2 units up.
We apply the following function transformation:
Vertical compressions:
To graph y = a * f (x)
If 0 <a <1, the graph of y = f (x) is compressed vertically by a factor a. (Shrinks)
We have then:
Horizontal translations:
Suppose that h> 0
To graph y = f (x + h), move the graph of h units to the left.
We have then:
Vertical translations:
Suppose that k> 0
To graph y = f (x) + k, move the graph of k units up.
We have then:
Answer:
Vertical compression by factor of 1/2. Horizontal displacement 3 units to the left. Vertical displacement 2 units up.
Answer : 2√3
<u>Given </u><u>:</u><u>-</u>
- A equilateral triangle with side length 4.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The value of x in the given figure.
As we know that in a equilateral triangle , perpendicular bisector , angle bisector and median coincide with each other .
- So the perpendicular drawn in the figure will bisect the given side .
- Therefore the value of each half will be 4/2 = 2 .
Now we may use Pythagoras theorem as ,
→ AB² = BC² + AC²
→ 4² = 2² + x²
→ 16 = 4 + x²
→ x² = 16-4
→ x² = 12
→ x =√12 = √{ 3 * 2²}
→ x = 2√3
<u>Hence </u><u>the</u><u> required</u><u> answer</u><u> is</u><u> </u><u>2</u><u>√</u><u>3</u><u> </u><u>.</u>
I hope this helps.
