Prove the theorem: In a plane, if a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the o
1 answer:
Answer:
The theorem requiring proof;
A transversal perpendicular to one of two parallel lines is also perpendicular to the other parallel line
The two column proof is as follows
Statement Reason
RS ⊥ CD Given
AB ║ CD Given
The Slope of AB = The slope of CD definition of parallel lines
The Slope of RS = -1/(The slope of CD) definition of perpendicular lines
The Slope of RS = -1/(The slope of AB) transitive property of equality
The Slope of RS = -1/(The slope of AB) definition of perpendicular lines
RS ⊥ AB
Step-by-step explanation:
You might be interested in
9514 1404 393
Explanation:
1) 2x^2 -5x -3 = 0 . . . . standard form equation
To convert this to factored form, you can look for factors of the product (2)(-3) that have a sum of -5. It can help to start by listing the ways that -6 can be factored. Since we want the sum of factors to be negative, we want to have larger negative factors.
-6 = (1)(-6) = (2)(-3)
The sums of these factor pairs are -5 (what we want) and -1 (not relevant). We can call these factors p=1 and q=-6.
If a = 2 is the leading coefficient of our standard form quadratic, we want to use these factors in the form ...
(ax +p)(ax +q)/a . . . . . factored form of the quadratic
(2x +1)(2x +(-6))/2 . . . .fill in the values we know
(2x +1)(x -3) . . . . . . . factor 2 from the second binomial
So, the factored form of the quadratic equation is ...
(2x +1)(x -3) = 0 . . . . factored form equation
__
2) f(x) = x^2 +7x +10 . . . . standard form quadratic function
Using the thinking process described above, we are looking for factors of 10 that have a sum of 7. We know those are 2 and 5. So, the factored form of the function is ...
f(x) = (x +2)(x +5) . . . . . . factored form quadratic function
The leading coefficient is 1, so we have no further work to do.
<u>Roots, x-intercepts, zeros</u>
The graph attached below shows this function crosses the x-axis when x=-2 and x = -5. These values of x are variously called "roots", "x-intercepts", and "zeros" of the function. They are values for which the factors and the function are zero. (x+2=0 when x=-2, for example)
<u>Solutions</u>
Often, we are interested in solving the equation ...
f(x) = 0
For that equation, the <em>solutions</em> <u>are</u> the <em>zeros</em> or <em>x-intercepts</em> or <em>roots</em>. The graph attached also shows solutions for ...
f(x) = 4
Those solutions are x = -6 and x = -1. The function value is not zero for these values of x, so the <em>roots</em>, <em>x-intercepts</em>, or <em>zeros</em> <u>are not</u> <em>solutions</em> to this equation.
