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masya89 [10]
3 years ago
11

Using the substitution method to solve the system. 5x+2y=5 x=3y-16

Mathematics
2 answers:
Aleksandr-060686 [28]3 years ago
8 0
5x+2y=5
x=3y-16
5(3y-16)+2y=5
15y-80+2y=5 collect the like terms
15y+2y=5+80
17y=85 divide both sides by 17
y=5

5x+2(5)=5
5x+10=5
5x=-5
x=-1
so x=-1 and y=5
seropon [69]3 years ago
6 0
5x + 2y = 5
Since, x = 3y - 16
Therefore,
5(3y - 16) + 2y = 5
15y - 80 +2y = 5
17y = 85
y = 5
therefore,
x = -1
y = 5
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A recipe for banana bread requires 3 cups of bananas for every 1 1/2 cups of sugar used.

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Naily [24]

Answers:

  • Satellite is approximately <u>2446.43 km</u> from station A.
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=========================================================

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

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

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h = x*tan(86.4)

We'll use this later in the substitution below.

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

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

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

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

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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))

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

We'll use this approximate x value to find h

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

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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.

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