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1 year ago

What is the solution(s) to the system of equations y = x² +4 and y = 2x + 4

Mathematics

1 answer:

bija089 [108]1 year ago

3 0

Answer:

$(x,y) = (0,4)~ \text{and}~ (x,y) = (2,8)$

Step by step explanation:

$y = x^2 +4~~~~~~~~~...(i)\\\\y = 2x +4~~~~~~~~~...(ii)\\\\\text{From (i) and (ii):}\\\\~~~~~~x^2 +4 = 2x+4\\\\\implies x^2 = 2x\\\\\implies x^2 -2x = 0\\\\\implies x(x-2) = 0\\\\\implies x =0,~ x = 2$

$\text{Substitute x = 0 in eq (i):}\\\\y = 0^2 +4 = 4\\\\\text{Substitute x = 2 in eq (i):}\\\\y=2^2+4 = 4+4 = 8\\\\\text{Hence,}~ (x,y)= \{(2,8), (0,4)\}$

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Step-by-step explanation:

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Answer:

The general form of the equation for the orbit of the satellite is $x^{2}+y^{2}+8\cdot x+2\cdot y = 4939.16$ .

Step-by-step explanation:

From statement we have the general equation of the elliptic form of Earth. First we have to transform this formula to the standard form of the equation of the circle. That is:

1) $x^{2}+y^{2}+8\cdot x +2\cdot y -4883=0$ Given

2) $(x^{2}+8\cdot x)+(y^{2}+2\cdot y) = 4883$ Associative, commutative and modulative properties/Compatibility with addition/Existence of additive inverse.

3) $(x^{2}+8\cdot x +16)+(y^{2}+2\cdot y+1)-17=4883$ Associative, commutative and modulative properties/Compatibility with addition/Existence of additive inverse.

4) $(x+4)^{2}+(y+1)^{2} = 4900$ Square perfect trinomial/Result

According the result, the center of Earth is $C(x,y) = (4, 1)$ and the square of the radius of Earth equals 4900. That is: $r = 70$

If satellite circles 0.4 unit above Earth, then the equation of the orbit of the satellite is:

$(x+4)^{2}+(y+1)^{2}=70.4^{2}$

$(x+4)^{2}+(y+1)^{2} = 4956.16$

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1) $(x+4)^{2}+(y+1)^{2} = 4956.16$ Given

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3) $x^{2}+y^{2}+8\cdot x+2\cdot y = 4939.16$ Associative, commutative and modulative properties/Compatibility with addition/Existence of additive inverse/Result

The general form of the equation for the orbit of the satellite is $x^{2}+y^{2}+8\cdot x+2\cdot y = 4939.16$ .

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