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3 years ago

What are the answers

Mathematics

1 answer:

Marizza181 [45]3 years ago

8 0

Answer:

Given any straight line and a point not on it, there "exists one and only one straight line which passes" through that point and never intersects the first line, no matter how far they are extended. This statement is equivalent to the fifth of Euclid's postulates, which Euclid himself avoided using until proposition 29 in the Elements. For centuries, many mathematicians believed that this statement was not a true postulate, but rather a theorem which could be derived from the first four of Euclid's postulates. (That part of geometry which could be derived using only postulates 1-4 came to be known as absolute geometry.)

Also draw the line straight line them up. To me it would be best if you use a ruler.

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What two numbers can add to 4 but multiple to -12

olasank [31]

Answer:

Numbers: 6 and -2

Step-by-step explanation:

Equations

This question can be solved by inspection. It's just a matter of factoring 12 into two factors that sum 4. Both numbers must be of different signs and they are 6 and -2. Their sum is indeed 6-2=4 and their product is 6*(-2)=-12.

However, we'll solve it by the use of equations. Let's call x and y to the numbers. They must comply:

$x+y=4\qquad\qquad [1]$

$x.y=-12\qquad\qquad [2]$

Solving [1] for y:

$y=4-x$

Substituting in [2]

$x(4-x)=-12$

Operating:

$4x-x^2=-12$

Rearranging:

$x^2-4x-12=0$

Solving with the quadratic formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

With a=1, b=-4, c=-12:

$\displaystyle x=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(-12)}}{2(1)}$

$\displaystyle x=\frac{4\pm \sqrt{16+48}}{2}$

$\displaystyle x=\frac{4\pm 8}{2}$

The solutions are:

$\displaystyle x=\frac{4+ 8}{2}=6$

$\displaystyle x=\frac{4- 8}{2}=-2$

This confirms the preliminary results.

Numbers: 6 and -2

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If this is your question, the answer is 4
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Answer:

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Step-by-step explanation:

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