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2 years ago

11

an object that weighs 1 pound on earth weighs 1.19 pounds on neptune suppose a dog weighs 9 pounds on earth how much would the s

ame dog weigh on neptune?

1 answer:

grigory [225]2 years ago

8 0

9x1.19=10.71

This means the dog weighs 10.71 pounds, hope this helps

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Answer:

30 degrees

Step-by-step explanation:

The angles must add up to 180 degrees and the triangles have the same angles.

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R = laps she runs
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WILL MARK BRAINLIEST WHOEVER ANSWERS CORRECTLY

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Answer:

Step-by-step explanation:

Let the relation between the total number of candies (y) and number of bags (x) is,

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3 years ago

Which equation has the solutions x=1+or-square root of 5?

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We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

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

$x^{2}+2x+1=-4+1$

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Rewrite as perfect squares

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therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

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

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

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

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

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

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

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