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kozerog [31]
3 years ago
9

PLEEEEEEEEEASE HEEEEEEEEELP

Mathematics
1 answer:
morpeh [17]3 years ago
7 0

Answer:

distance island dock to Dock A = 4.99 km

distance island dock to Dock K = 6.35 km

Step-by-step explanation:

Always make a scetch to visualize the situation.

You need to construct two triangle both with a streight angle, so you can use Pythagoras to calculate the unknown distances between the island dock L, and each of the other two docks A an K.

I chose to introduce an extra letter, the letter C. In total you have the letters A K L and the letter C.

The letter C has a streight angle of 90° between ACL and it has the same streight angle of 90° with KCL. It is crucial that you see that the distance of LC is exactly the same in triangle LAC and that LC has exactly the same distance in the other triangleLKC.

The distance between AK = 2.3 km.

I define the distance between K and point C as 2.3 + x, because the distance x is unknown.

KC = 2.3 + x

Further more, when you make a picture, you can see that the distance between A and point C = x.

From such a picture, it would show clearly, that K is further away in respect to L then point A. From the picture it would be clear that the angle of LKC is smaller then the angle of LAC, so LKC = 45° and LAC = 64°.

Because angle LKC = 45° and we choose C to have an angle of 90°, the TRIANGLE LKC must be a special triangle... In any triangle, the sum of the three angles together, must add up to 180° .

If that is true, then we have 45 + 90 + 45 (because that adds up to 180). Now that means triangle LKC must have two equal sides (because of the same angels of 45° ).

So we know the distance KC = LC and we already defined KC = 2.3 + x.

Now we know enough to solve the problem.

AK = 2.3 km

angle of LKC = 45°

angle of LAC = 64°

AC = x

KC = 2.3 + x

LC = KC

LC = 2.3 + x

Try to calculate the distance x by using tan. After that you can use Pythagoras to find the other distances.

tan(LKC) = ( LC ) / ( KC )

tan(LKC) = ( x+2.3 ) / ( x+2.3 )

That is not helpful. Let's try the other triangle...

tan(LAC) = LC / AC

tan(LAC) = ( x+2.3 ) / x

tan(64) = ( x+2.3 ) / x

Solve the equation which means you try to find the value for x.

x * tan(64) = ( x+2.3 )

tan(64) * x -x = 2.3

tan(64) * x - 1* x = 2.3

Try to get x outside of the braquets...

x* ( tan(64) - 1 ) = 2.3

x* (2.0503038415793 - 1 ) = 2.3

1.0503038415793 * x = 2.3

x = 2.3 / 1.0503038415793

x = 2.19

Now use Pythagoras a² + b² = c² in triangle LAC to find distance LA.

LA² = AC² + LC²

AC = x = 2.19

LC = 2.3 + x = 4.39

LA² = 2.19² + 4.39²

LA = SQRT( 4.79 + 20.16 )

LA = SQRT( 24.95 )

LA = 4.99 km

Now use Pythagoras a² + b² = c² in triangle LKC to find distance LK.

LK² = KC² + LC²

KC = 2.3 + x = 4.39

LC = 2.3 + x = 4.39

LK² = 4.39² + 4.39²

LK = SQRT( 20.16 + 20.16 )

LK = SQRT( 40.32 )

LK = 6.35 km

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Help me on these questions
mojhsa [17]

Answer:

a) The equation is (y - 1)² = -8 (x - 4)

b) The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

Step-by-step explanation:

a) Lets revise the standard form of the equation of the parabola with a

   horizontal axis

# (y - k)² = 4p (x - h), (h , k) are the coordinates of its vertex and p ≠ 0

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* Lets solve the problem

∵ The focus is (2 , 1)

∵ focus is (h + p , k)

∴ h + p = 2 ⇒ subtract p from both sides

∴ h = 2 - p ⇒ (1)

∴ k = 1

∵ It opens left, then the axis is horizontal and p is negative

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∵ k = 1

∴ Its equation is (y - 1)² = 4p (x - h)

- The parabola contains point (2 , 5), substitute the coordinates of the

 point in the equation of the parabola

∴ (5 - 1)² = 4p (2 - h)

∴ (4)² = 4p (2 - h)

∴ 16 = 4p (2 - h) ⇒ divide both sides by 4

∴ 4 = p (2 - h) ⇒ (2)

- Use equation (1) to substitute h in equation (2)

∴ 4 = p (2 - [2 - p]) ⇒ open the inside bracket

∴ 4 = p (2 - 2 + p) ⇒ simplify

∴ 4 = p (p)

∴ 4 = p² ⇒ take √ for both sides

∴ p = ± 2, we will chose p = -2 because the parabola opens left

- Substitute the value of p in (1) to find h

∵ h = 2 - p

∵ p = -2

∴ h = 2 - (-2) = 2 + 2 = 4

∴ The equation of the parabola in standard form is

  (y - 1)² = 4(-2) (x - 4)

∴ The equation is (y - 1)² = -8 (x - 4)

b) Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with  center (h , k)

 and major axis parallel to x-axis is (x - h)²/a² + (y - k)²/b² = 1  

- The coordinates of the vertices are (h ± a , k )  

- The coordinates of the foci are (h ± c , k), where c² = a² - b²  

* Now lets solve the problem

∵ Its vertices are (-4 , 4) and (6 , 4)

∵ The coordinates of the vertices are (h + a , k ) and (h - a , k)  

∴ k = 4

∴ h + a = 6 ⇒ (1)

∴ h - a = -4 ⇒ (2)

- Add (1) and (2) to find h

∴ 2h = 2 ⇒ divide both sides by 2

∴ h = 1

- Substitute the value of h in (1) or (2) to find a

∴ 1 + a = 6 ⇒subtract 1 from both sides

∴ a = 5

∵ The foci at (-2 , 4) and (4 , 4)

∵ The coordinates of the foci are (h + c , k) , (h - c , k)

∴ h + c = 4

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∴ c = 3

∵ c² = a² - b²

∴ 3² = 5² - b²

∴ 9 = 25 - b² ⇒ subtract 25 from both sides

∴ -16 = -b² ⇒ multiply both sides by -1

∴ 16 = b²

∵ a² = 25

∵ The equation of the ellipse is (x - h)²/a² + (y - k)²/b² = 1

∴ The equation is (x - 1)²/25 + (y - 4)²/16 = 1

c) How to identify the type of the conic  

- Rewrite the equation in the general form,  

 Ax² + Bxy + Cy² + Dx + Ey + F = 0  

- Identify the values of A and C from the general form.  

- If A and C are nonzero, have the same sign, and are not equal  

 to each other, then the graph is an ellipse.  

- If A and C are equal and nonzero and have the same sign, then

 the graph is a circle  

- If A and C are nonzero and have opposite signs, and are not equal  

 then the graph is a hyperbola.  

- If either A or C is zero, then the graph is a parabola  

* Now lets solve the problem

∵ x² + 4y² - 6x - 7 = 0

∵ The general form of the conic equation is

   Ax² + Bxy + Cy² + Dx + Ey + F = 0  

∴ A = 1 and C = 4

∵ If A and C are nonzero, have the same sign, and are not equal  to

  each other, then the graph is an ellipse.

∵ x² + 4y² - 6x - 7 = 0 ⇒ re-arrange the terms

∴ (x² - 6x ) + 4y² - 7 = 0

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∵ 6x ÷ 2 = 3x

∵ 3x = x × 3

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 x² - 6x + 9 = (x - 3)²

∴ (x² - 6x + 9) - 9 + 4y² - 7 = 0 ⇒ simplify

∴ (x - 3)² + 4y² - 16 = 0 ⇒ add 16 to both sides

∴ (x - 3)² + 4y² = 16 ⇒ divide all terms by 16

∴ (x - 3)²/16 + 4y²/16 = 1 ⇒ simplify

∴ (x - 3)²/16 + y²/4 = 1

∴ The equation of the ellipse is (x - 3)²/16 + y²/4 = 1

5 0
3 years ago
A bag contains 5 red poker chips, 3 blue poker chips and 7 white poker chips. Two chips are drawn without replacement. Find the
prohojiy [21]
5 red, 3 blue, 7 white.....total of 15

probability of 1st drawn being red : 5/15 reduces to 1/3
without replacement
probability of 2nd being red : 4/14 reduces to 2/7

probability of both : 1/3 * 2/7 = 2/21 <==
4 0
3 years ago
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