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- Find x2z ds, where C is the line segment from (0,-3,2) to (-3,-2,8)Which of the points P = (-1,4,-7) and Q = (6,-1,5) is closest to the z-axis? OQ (6,-1,5) OQ (6,-1,0) OP=(-1,4,-7) OP=(-1,4,0)1.- A circle passes through the points (2,1) and (-2,3) and its center is on the line 2x+y+4=0. Find its equation.
- The triangle whose vertices are (2,5),(3,1),(2,5),(3,1), and (4,2)(4,2) is transformed by the rule (x,y)→(x−2,y+4)(�,�)→(�−2,�+4). Is the image similar or congruent to the original figure?the vertices A(-2,-1), B(-3,2), C(-1,3), and D(0,0) form a prallelogram. The vertices A'(-1,-2), B'(2,-3), C'(3,-1) and D's(0,0) are the image of the parallelogram after a sequence of transformations. which sequence of transformations could produce the image from the pre-image?Find the set of values of 'a' if the points (1, 2) and (3, 4) lie to the same side of the straight line 3x -5y + a= 0.
- Determine the new order pairs if the points are reflected across the y- axis. J(2, 2), K(7, 4), L(9, -2), M(3, -1) O J (2.2), K (4,7),L (2,9), M'(1,3) O '(2,2), K (-4,7), E (2,-9), M (1,-3) O s(22). K(7,4), L(-9, -2), M(-3, -1)6. For each equation in Question 1.a), replace x with 0 to find y. If you don't get the y-intercept, find your mistake. i) y= 2x – 1 ii) y= –1.5x+ 2 iii) y = -x - 0.5 = 2(0) – 1 = -1/ 1 iv) y=-X - 3 1 v) y =-2x+ vi) ! X- 112 1/2Which of the following points (x,y) is NOT on the graph of y = 1/(x2 + 1)? a. (3, 1/10) b. (2, 1/5) c. (0,1) d. (1,2)
- and K(-4, 1). 4 a 3- 2 -6 -5 -4 -3 -2 -1 -2 -3 -4 -6. 6. A pre-image has coordinates A(3, -5), B(3, 1) and C(-2, 0). The image has coordinates A'(3, 5), B’(3, –1) and C’(-2, 0). What type of reflection occurred? Explain how you know. " The point (7, 0) is reflected over the x-axis. What do you notice about the coordinates of the reflected point? Why is this? Lesson 7.1- ReflectionsWhat are the vertices of ΔA'B'C produced by T−3, 6 (ΔABC) = ΔA'B'C? A. A′(0, 6), B′(0, 4), C′(−3, 3) B. A′(6, 6), B′(6, 4), C′(3, 3) C. A′(0, −6), B′(0, −8), C′(−3, 9) D. A′(6, −6), B′(6, −8), C′(3, 9)What is the symmetric equation of the line through the points (2,1,11) and (5,-1,2). O a. (1-2x)/2)=((2y-1)/2)=(1+2z) O b. ((2-x)/3)=((y-1)/2)=(1-z) O c. (3-2x)=(2y-3)%3D((2z-1)/2) O d. (1-x)=((y+1)/3)=((1-Z)/2) O e. ((2-2x)/3)=((2y+1)/-2)=(2z)