A and E is 8 while Q and V are 12. The scale factor will help us find M. 8 goes into 12 1 1/2 times. If you multiply 6 by 1 1/12 you get M.
Answer: 9
Answer:
x = ± 4
Step-by-step explanation:
Given
x² = 96 ( take the square root of both sides )
x = ± 
= ± 
= 
× = ± 4
The correct answer is " (1) Reflect ABC across the x-axis and call this new triangle A'B'C'. (2) Translate A'B'C' 2 units right and 6 units up so that its image is A''B''C''. "
It is assumed that the points are
A in ABC is (1,9), B in ABC is (3,12), and C in ABC is (4,4).
A'' in A''B''C'' is (3,-3), B'' in A''B''C'' is (5,-6), and C'' in A''B''C'' is (6,2).
Both triangles are congruent. Since they are congruent, there are no contractions nor dilations occured. After rotating clockwise, we get A"C"B". So we need to reflect it to get A"B"C"
hope this helped :D
<span><span> <span>Akar akar persamaan kuadrat 2x² - 3x -1 = 0 adalah x1 dan x2. Persamaan kuadrat baru yang akar akarnya satu lebih kecil dari dua kali akar akar persamaan kuadrat di atas adalah ........</span></span><span><span><span>A.x² - x - 4 = 0</span><span>B.x² + 5x - 4 = 0</span><span>C.x² - x + 4 = 0</span></span><span><span>D.x² + x + 4 = 0</span><span>E.x² - 5x - 4 = 0</span></span></span><span>Jawaban : A
Penyelesaian :
Akar-akar persamaan lama : x1 dan x2
Akar-akar persamaan baru : xA dan xB
xA = 2x1 - 1
xB = 2x2 - 1
xA + xB = (2x1 - 1) + (2x2 - 1)
= 2 (x1 + x2) - 2
= 2 () - 2
= 3 - 2
xA + xB = 1
xA . xB = (2x1 - 1) (2x2 - 1)
= 4 x1.x2 - 2(x1 + x2) + 1
= 4.(-) - 2() + 1
= -2 - 3 + 1
xA . xB = -4
Jadi persamaan kuadrat baru : x² - (xA + xB)x + xA . xB = 0
x² - x - 4 = 0
</span></span>
