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meriva
3 years ago
12

2(x+y) = 61-2y 4x-4 = 18-3y

Mathematics
1 answer:
Naddik [55]3 years ago
4 0

Answer:

x = 8

y = -7

Step-by-step explanation:

This is a system of equations called simultaneous equations. We shall solve it by elimination method Step 1We shall label the equations (1) and (2)−3y−4x=−11.....(1)3y−5x=−61......(2)Step 2Multiply each term in equation (1) by 1 to give equation (3)1(-3y-4x=-11).....(1)-3y-4x=-11....(3)Step 3Multiply each term in equation 2 by -1 to give equation (4)-1(3y−5x=−61)......(2)-3y+5x=61.....(4)Step 4-3y-4x=-11....(3)-3y+5x=61.....(4)Subtract each term in equation (3) from each term in equation (4)-3y-(-3y)+5x-(-4x)=61-(-11)-3y+3y+5x+4x=61+110+9x=729x=72Step 5Divide both sides of the equation by 9, the coefficient of the unknown variable x to find the value of x 9x/9 = 72/9x = 8Step 6Put in x = 8 into equation (2)3y−5x=−61......(2)3y-5(8)=-613y-40=-61Collect like terms by adding 40 to both sides of the equation 3y-40+40=-61+403y=-21Divide both sides by 3, the coefficient of y to find the value of y 3y/3=-21/3y=-7Therefore, the values of x and y that satisfy the equations are 8 and -7 respectively

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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
3 years ago
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Answer:

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Step-by-step explanation:

24 * 2 = 48

208 - 48 = 160

160 / 16 = 10

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Answer:

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Step-by-step explanation:

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