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Firdavs [7]
2 years ago
6

Solve the system of linear equations by graphing. Round the solution to the nearest tenth as needed. Y + 2.3 = 0.45x and -2y = 4

.2x - 7.8
Mathematics
2 answers:
Setler [38]2 years ago
5 0

Answer:

a

Step-by-step explanation:

HACTEHA [7]2 years ago
4 0

Answer:

(2.4, -1.2)

Step-by-step explanation:

Start by moving the x and the y to the same side and moving the number across the equal sign in both equations. We should now have y-0.45x=-2.3 and 2y+4.2x=7.8. We can use the elimination method by multiplying the first equation by -2 to get -2y+0.9x=4.6 and 2y+4.2x=7.8. From there, add the two equations together, eliminating y (-2+2=0). We now have 5.1x=12.4; divide both sides by 5.1 to get x=2.4. Then, in any of the two equations, let's use y-0.45x=-2.3, substitute x with 2.4. Now we have y-1.08=-2.3. Add 1.08 to both sides to get y=-1.22; round that to the nearest tenth to get -1.2.

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we have

7=-2x^{2} +10x

Factor the leading coefficient

7=-2(x^{2} -5x)

Complete the square. Remember to balance the equation by adding the same constants to each side

7-12.50=-2(x^{2} -5x+2.5^{2})

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Divide both sides by -2

2.75=(x^{2} -5x+2.5^{2})

Rewrite as perfect squares

2.75=(x-2.5)^{2}

Taking the square roots of both sides (square root property of equality)

x-2.5=(+/-)\sqrt{2.75}

Remember that

\sqrt{2.75}=\sqrt{\frac{11}{4}}= \frac{\sqrt{11}}{2}

x-2.5=(+/-)\frac{\sqrt{11}}{2}

x=2.5(+/-)\frac{\sqrt{11}}{2}

x=2.5+\frac{\sqrt{11}}{2}=\frac{5+\sqrt{11}}{2}

x=2.5-\frac{\sqrt{11}}{2}=\frac{5-\sqrt{11}}{2}

<u>the answer is</u>

The solutions are

x=\frac{5+\sqrt{11}}{2}

x=\frac{5-\sqrt{11}}{2}


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Consider the initial value problem y′+4y=48t,y(0)=9. y′+4y=48t,y(0)=9. Take the Laplace transform of both sides of the given dif
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Answer:

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Step-by-step explanation:

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