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Bas_tet [7]
3 years ago
11

A fighter pilot needs to make an emergency landing. He needs a clear space that is at least 600 m long to land safely. He is fly

ing at an altitude of 300 m. He sees a small empty field. The angles of depression from the jet to the two ends of the field are 30° and 56° [5]
a) How long is the field?
b) Can he land safely?
Mathematics
1 answer:
Zolol [24]3 years ago
4 0

(a) See the attached sketch. The jet makes a right triangle with the ground with hypotenuse <em>y</em>. Since the angle of depression is 56°, and the plane in which the jet is flying 300 m in the air is parallel with the ground, the angle of elevation in the right triangle is also 56°. Use the definition of sine to solve for <em>y</em>.

sin(56°) = (300 m)/<em>y</em>   →   <em>y</em> = (300 m)/sin(56°)

The uppermost angle in the blue triangle has measure equal to the difference between the two given angles of depression, 56° - 30° = 26°, and the rightmost angle is also 30° by the same reasoning as before. Then by the law of sines,

<em>x</em>/sin(26°) = <em>y</em>/sin(30°)   →   <em>x</em> = sin(26°)/sin(30°) <em>y</em> ≈ 317.2627 m

(b) No, the field is not long enough.

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Leni [432]

Answer:

-2, -5

Step-by-step explanation:

h2 + 7h + 10 = 0

h^2 +7h + 10 = 0

h^2 +2h +5h + 10 = 0

h(h+2) + 5(h+2) = 0

h+2 = 0 or h+5 = 0

h = -2 or h = -5

4 0
3 years ago
What are three collinear points in the figure below?
algol [13]

Answer:

B. W,V,Z

Step-by-step explanation:

Collinear points are two more points in the same plane that lie in a straight line.

From the diagram, Z is the intersection of the two diagonals of quadrilateral VXWY.

The points V, Z and W lie along the same diagonal, hence they are collinear.

The point y, Z, and X also lie along the same diagonal, these are also collinear.

The correct choice is B

7 0
3 years ago
4. Evaluate the expression for the given value of each variable.
Novosadov [1.4K]

Answer:

-18

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5a2-7-b3

5*a*2-7-b*3

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Hope this helped! :)

5 0
3 years ago
Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio
JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

Initial =1\ million

6\ years\ later = 500,000

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

\frac{dA}{dt} = kA

Where k represents the constant of proportionality

\frac{dA}{dt} = kA

Multiply both sides by dt/A

\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}

\frac{dA}{A}  = k\ dt

Integrate both sides

\int\ {\frac{dA}{A}  = \int\ {k\ dt}

ln\ A = kt + lnC

Make A, the subject

A = Ce^{kt}

t = 0\ when\ A =1\ million i.e. At initial

So, we have:

A = Ce^{kt}

1000000 = Ce^{k*0}

1000000 = Ce^{0}

1000000 = C*1

1000000 = C

C =1000000

Substitute C =1000000 in A = Ce^{kt}

A = 1000000e^{kt}

To solve for k;

6\ years\ later = 500,000

i.e.

t = 6\ A = 500000

So:

500000= 1000000e^{k*6}

Divide both sides by 1000000

0.5= e^{k*6}

Take natural logarithm (ln) of both sides

ln(0.5) = ln(e^{k*6})

ln(0.5) = k*6

Solve for k

k = \frac{ln(0.5)}{6}

k = \frac{-0.693}{6}

k = -0.1155

Recall that:

\frac{dA}{dt} = kA

Where

\frac{dA}{dt} = Rate

So, when

A = 600000

The rate is:

\frac{dA}{dt} = -0.1155 * 600000

\frac{dA}{dt} = -69300

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0
3 years ago
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vaieri [72.5K]

Answer:

\frac{4}{5}

Step-by-step explanation:

Given

sinΘ = \frac{3}{5} = \frac{opposite}{hypotenuse}

Then the hypotenuse = 5 and one leg of the right triangle is 3

Using Pythagoras' identity to find the second leg (x)

x² + 3² = 5²

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x² = 16 ⇒ x = 4

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7 0
3 years ago
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