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Gnom [1K]
3 years ago
7

A plane has traveled 26 miles on a course heading 48º east of north. How far north (x) has the plane traveled at this point, to

the nearest tenth of a mile?
Mathematics
1 answer:
GrogVix [38]3 years ago
3 0

Answer: 17.4\ \text{miles}

Step-by-step explanation:

Given

A plane has traveled 26 miles on a course heading 48^{\circ} east of north

Taking the component of travel in North direction

The distance traveled in North is given by with the help of figure

\Rightarrow x=26\cos 48^{\circ}\\\\\Rightarrow x=26\times 0.669\\\\\Rightarrow x=17.4\ \text{miles}

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Find the reciprocal
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Answer:

Depends on what number(s)/fraction(s) your working with.

Step-by-step explanation:

Find the reciprocal of a fraction by flipping it. The definition of "reciprocal" is simple. To find the reciprocal of any number, just calculate "1 ÷ (that number)." For a fraction, the reciprocal is just a different fraction, with the numbers "flipped" upside down (inverted).

3 0
3 years ago
Let x = 4.<br><br><br><br> What is the value of y in the equation y = 15 + 11 + x?
marishachu [46]
Okay so you have to add up 15 and 11 and 4 so your answer is y=30


5 0
3 years ago
8. Aimi decided to spend her money and save some in the ratio of 3:2. If her total money is RM
Bogdan [553]

Step-by-step explanation:

2/5×7500

2×1500

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3 0
3 years ago
the eccentricity of a hyperbola is defined as e=c/a. find an equation with vertices (1,-3) and (-3,-3) and e=5/2
Ksivusya [100]

Answer:

\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1.

Step-by-step explanation:

If (α, β) are the coordinates of the center of the hyperbola, then its equation of the hyperbola is \frac{(x - \alpha )^{2} }{a^{2} } - \frac{(y - \beta )^{2} }{b^{2} } = 1.

Now, the vertices of the hyperbola are given by (α ± a, β) ≡ (1,-3) and (-3,-3)

Hence, β = - 3 and α + a = 1 and α - a = -3

Now, solving those two equations of α and a we get,  

2α = - 2, ⇒ α = -1 and

a = 1 - α = 2.

Now, eccentricity of the hyperbola is given by b^{2} = a^{2}(e^{2}  - 1) = 4[(\frac{5}{2} )^{2} -1] = 21 {Since e = \frac{5}{2} given}

Therefore, the equation of the given hyperbola will be  

\frac{(x + 1 )^{2} }{4} - \frac{(y + 3 )^{2} }{21} = 1. (Answer)

6 0
3 years ago
Olve for x. 3(3x - 1) + 2(3 - x) = 0 x = 3/7 x = -3/7 x = 1/3 x = -1/3
zloy xaker [14]

<u>Answer:</u>

x = -3/7

<u>Step-by-step explanation:</u>

We are given the following equation for which we have to solve for x:

3(3x - 1) + 2(3 - x) = 0

So we will start by expanding the equation by multiplying the common factor with the brackets to get:

9x-3+6-2x=0

Arranging the like terms together (variables on one side of the equation and constants on the other side) to get:

9x-2x=3-6\\\\7x=-3\\\\x=-\frac{3}{7}

Therefore, x = -3/7.

8 0
4 years ago
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