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antoniya [11.8K]
2 years ago
7

Can you please help me please ​

Mathematics
1 answer:
lianna [129]2 years ago
7 0
On the y said it would be -2
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Under what circumstances does the system of equations Qx+Ry=S and Y=Tx+S have infinitely many solutions?
Anettt [7]
From these -Tx+y=S. If -T=Q/R, then y=-Qx/R+S, so Ry=-Qx+RS, Qx+Ry=RS=S.
If R is not equal to 1, or S is non-zero, the equations are inconsistent, so there would be no solutions.
If R=1 there are an infinite number of solutions given by Qx+y=S, or y=S-Qx or y=S+Tx.
If S=0, Qx+Ry=0 or y=-Qx/R or y=Tx.
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3 years ago
5) (5x2 + 4) — (5 + 5x3)
Licemer1 [7]
(5x²+4)–(5+5x³)
25x+4–5+125x
150x–1
8 0
3 years ago
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
What is .015 as fraction
schepotkina [342]
It is 15 / 1000 in the fraction form.
4 0
2 years ago
Read 2 more answers
A wheel has a radius of 15 cm. Approximately how far does it travel in 4 revolutions?
ArbitrLikvidat [17]
Radius is 15 cm
Circumference = 2 pi r
C = 2 x 3.14 x 15cm
C = 94.2 cm
94.2 cm x 4 (revolutions) = 376.8 = D
7 0
3 years ago
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