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aleksandr82 [10.1K]
3 years ago
13

Question 11:

Mathematics
1 answer:
den301095 [7]3 years ago
8 0

Answer:

69 miles per hour

Step-by-step explanation:

Given

Trip to:

t_1 = 6 --- time

s_1 = 23 + s_2

Trip back

t_2 = 9

Required

Determine the average speed to Johannesburg

Average speed is calculated as:

speed = \frac{distance}{time}

Make distance the subject

distance = speed * time

Since distance is constant, then:

s_1 * t_1 = s_2 * t_2

This gives:

(23 + s_2) * 6 = s_2 * 9

Open bracket

138 + 6s_2 = 9s_2

Collect like terms

9s_2 - 6s_2 = 138

3s_2 = 138

Solve for s2

s_2 = 138/3

s_2 = 46

Recall that:

s_1 = 23 + s_2

s_1 = 23 + 46

s_1 = 69

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Answer:

Anne’s after-tax rate of return from the corporate bond is 3.5% or 5% x (1-.3). Because interest from the bond is taxed annually and her rate is assumed to be constant, the after-tax rate of return doesn’t depend on her investment horizon. Thus, her annual after-tax rate of return remains at 3.5% if the bond matures in ten years.

Step-by-step explanation:

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3 years ago
A bar graph is used to represent numerical data.<br><br> true or false
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4) The path of a satellite orbiting the earth causes it to pass directly over two
Naily [24]

Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
  • Satellite is approximately <u>2441.61 km</u> above the ground.

=========================================================

Explanation:

I'm assuming tracking stations A and B are at the same elevation and are on flat ground. In reality, this is likely not the case; however, for the sake of simplicity, we'll assume this is the case.

The diagram is shown below. Points A and B describe the two stations, while point C is the satellite's location. Point D is on the ground directly below the satellite. We have these lengths

  • AB = 60 km
  • AD = x
  • CD = h

Focusing on triangle ACD, we can apply the tangent rule to isolate h.

tan(angle) = opposite/adjacent

tan(A) = CD/AD

tan(86.4) = h/x

x*tan(86.4) = h

h = x*tan(86.4)

We'll use this later in the substitution below.

--------------------

Now move onto triangle BCD. For the reference angle B = 85, we can use the tangent rule to say

tan(angle) = opposite/adjacent

tan(B) = CD/DB

tan(B) = CD/(DA+AB)

tan(85) = h/(x+60)

tan(85)*(x+60) = h

tan(85)*(x+60) = x*tan(86.4) .............  apply substitution; isolate x

x*tan(85)+60*tan(85) = x*tan(86.4)

60*tan(85) = x*tan(86.4)-x*tan(85)

60*tan(85) = x*(tan(86.4)-tan(85))

x*(tan(86.4)-tan(85)) = 60*tan(85)

x = 60*tan(85)/(tan(86.4)-tan(85))

x = 153.612786190499

--------------------

We'll use this approximate x value to find h

h = x*tan(86.4)

h = 153.612786190499*tan(86.4)

h = 2441.60531869599

h = 2441.61 km  is how high the satellite is above the ground.

Return to triangle ACD. We'll use the cosine rule to determine the length of the hypotenuse AC

cos(angle) = adjacent/hypotenuse

cos(A) = AD/AC

cos(86.4) = x/AC

cos(86.4) = 153.612786190499/AC

AC*cos(86.4) = 153.612786190499

AC = 153.612786190499/cos(86.4)

AC = 2446.43279498247

AC = 2446.43 km is the distance from the satellite to station A.

6 0
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Answer:

x = 27

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180 = 5x + 2 + x + 16

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180 - 18 = 6x

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162/6 = x

27 = x

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