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12345 [234]
3 years ago
11

To approach the runway, a pilot of a small plane must begin at 6 degrees,descent starting from a height of 1905 feet above the g

round. To the nearest tenth of a mile, how many miles from the runway is the airplane at the start of this approach

Mathematics
1 answer:
koban [17]3 years ago
6 0

Answer:

C. 3.5 mi

Step-by-step explanation:

Reference angle = angle of elevation

Angle of elevation = angle of depression = 6° (alternate angle theorem)

Hypotenuse = x

Opposite = 1905 ft

Apply trigonometric function SOH:

Sin 6° = Opp/Hyp

Sin 6° = 1905/x

x * Sin 6° = 1905

x = 1905/Sin 6°

x = 18,224.7 ft

Convert form feet to miles

1 mi = 5,280 ft

Therefore,

18,224.7 = 18,224.7/5,280

= 3.45164773 mi

≈ 3.5 mi (nearest mile)

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£6000 is divided between Adam, Ben and<br> Chris in the ratio 1:3:4. How much does Ben receive?
Gala2k [10]

Answer:

Chris gets £3000

Step-by-step explanation:

A : B : C

1 : 3 : 4 = 8 (add all the parts together)

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2 years ago
Consider the following quadratic function Part 3 of 6: Find the x-intercepts. Express it in ordered pairs.Part 4 of 6: Find the
maksim [4K]

Answer:

The line of symmetry is x = -3

Explanation:

Given a quadratic function, we know that the graph is a parabola. The general form of a parabola is:

y=ax^2+bx+c

The line of symmetry coincides with the x-axis of the vertex. To find the x-coordinate of the vertex, we can use the formula:

x_v=-\frac{b}{2a}

In this problem, we have:

y=-x^2-6x-13

Then:

a = -1

b = -6

We write now:

x_v=-\frac{-6}{2(-1)}=-\frac{-6}{-2}=-\frac{6}{2}=-3

Part 3:

For this part, we need to find the x-intercepts. This is, when y = 0:

-x^2-6x-13=0

To solve this, we can use the quadratic formula:

x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot(-1)\cdot(-13)}}{2(-1)}

And solve:

x_{1,2}=\frac{6\pm\sqrt{36-52}}{-2}x_{1,2}=\frac{-6\pm\sqrt{-16}}{2}

Since there is no solution to the square root of a negative number, the function does not have any x-intercept. The correct option is ZERO x-intercepts.

Part 4:

To find the y intercept, we need to find the value of y when x = 0:

y=-0^2-6\cdot0-13=-13

The y-intercept is at (0, -13)

Part 5:

Now we need to find two points in the parabola. Let-s evaluate the function when x = 1 and x = -1:

x=1\Rightarrow y=-1^2-6\cdot1-13=-1-6-13=-20x=-1\Rightarrow y=-(-1)^2-6\cdot(-1)-13=-1+6-13=-8

The two points are:

(1, -20)

(-1, -8)

Part 6:

Now, we can use 3 points to find the graph of the parabola.

We can locate (1, -20) and (-1, -8)

The third could be the vertex. We need to find the y-coordinate of the vertex. We know that the x-coordinate of the vertex is x = -3

Then, y-coordinate of the vertex is:

y=-(-3)^2-6(-3)-13=-9+18-13=-4

The third point we can use is (-3, -4)

Now we can locate them in the cartesian plane:

And that's enough to get the full graph:

8 0
1 year ago
Plz help me well mark brainliest if correct!!...
valina [46]

Answer:

You should stick with option D

4 0
3 years ago
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