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const2013 [10]
3 years ago
15

One end point of a line segment is ( -3, -6 ) The length of the line segment is 7 units find four points that could serve as the

other end point of the given line segment
Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

Answer:

<em>(-3, 1), (-3, -13), (4, -6), (-10, -6)</em>

Step-by-step explanation:

You can think of point (-3, -6) as the center of a circle of radius 7. Then all points on the circle are 7 units from point (-3, -6). There are 4 points 7 units away from point (-6, -3) whose coordinates are easy to find. Think of a vertical line and a horizontal line that go through point (-3, -6). Up and sown from point (-3, -6) are points (-3, 1) and (-3, -13). To the right and left of point (-3, -6) are points (4, -6) and (-10, -6).

Mrac [35]3 years ago
6 0

Hey lassie. Here si your answer for this question.

Answer: (4,-6), (-3,-13), (-10,-6), and (-3,1).

Hope this helps you.

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What is this fraction in its simplest form?<br>​
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a^2/3

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

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