3 lines are shown. A line with points T, R, W intersects with a line with points S, R, V at point R. Another line extends from p
oint R to point U in between angle V R W. Which pair of angles are vertical angles? AngleWRU and AngleSRT AngleWRS and AngleVRT AngleVRU and AngleTRS AngleVRT and AngleSRT
2 answers:
Answer:
The vertical angles are ∠ WRS and ∠ VRT ⇒ 2nd answer
Step-by-step explanation:
* <em>Lets explain the meaning of the vertical angles</em>
- Vertical angles are the angles opposite to each other when two lines
are intersected
- The vertical angle are equal in measures
- <em>Look to attached figure which represents the lines </em>
∵ Line TW intersects line SV at R
∴ ∠ WRV and ∠ SRT are vertical angle
∴ <u><em>∠ WRS and ∠ VRT</em></u> are vertical angle
∵ RU is between the rays of angle VRW
∴ RU intersects the lines TW and SV at R
∴ No vertical angles between RU and the other two lines because
it intersects them its endpoint R
* <em>Now lets look to the answer to chose the right one</em>
- ∠ WRU and ∠ SRT are not vertical angles
- <em><u>∠ WRS and ∠ VRT are vertical angles </u></em>
- ∠ VRU and ∠ TRS are not vertical angles
- ∠ VRT and ∠ SRT are not vertical angles
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Answer:
(D)The midpoint of both diagonals is (4 and one-half, 5 and one-half), the slope of RP is 7, and the slope of SQ is Negative one-sevenths.
Step-by-step explanation:
- Point P is at (4, 2),
- Point Q is at (8, 5),
- Point R is at (5, 9), and
- Point S is at (1, 6)
Midpoint of SQ
Midpoint of PR
Now, we have established that the midpoints (point of bisection) are at the same point.
Two lines are perpendicular if the slope of one is the negative reciprocal of the other.
In option D
- Slope of RP =7
- Slope of SQ
Therefore, lines RP and SQ are perpendicular.
Option D is the correct option.
<h3>Answer:</h3>
- f(x) = -2x^3 +3x^2 +11x -6
- see attached
- an infinite number. Since the magnitude of the leading coefficient is not specified, it may be any negative number. (We have chosen the smallest magnitude integer that makes all coefficients be integers.)
1. When "a" is a root of a polynomial, (x -a) is a factor of it. For the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).
In order to make the leading coefficient be negative, we need to multiply this product by a negative number. Any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.
Then ...
... f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)
... = -(2x² +3x -2)(x -3)
... = -(2x³ -3x² -11x +6)
... f(x) = -2x³ +3x² +11x -6
2. A graph created by the Desmos on-line graphing calculator is shown, and the zeros are highlighted.
3. As indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. You want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.