Answer: The correct option is (D) P″(9, -12) and Q″(15, -3).
Step-by-step explanation: Given that triangle PQR is dilated by a scale factor of 1.5 to form triangle P′Q′R′. This triangle is then dilated by a scale factor of 2 to form triangle P″Q″R″.
The co-ordinates of vertices P and Q are (3, -4) and (5, -1) respectively.
We are to find the co-ordinates of the vertices P″ and Q″.
<u>Case I :</u> ΔPQR dilated to ΔP'Q'R'
The co-ordinates of P' and Q' are given by
<u>Case II :</u> ΔP'Q'R' dilated to ΔP''Q''R''
The co-ordinates of P'' and Q'' are given by
Thus, the co-ordinates of the vertices P'' and Q'' are (9, -12) and (15, -3).
Option (D) is CORRECT.
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Answer:903.55
Step-by-step explanation:
At the time the rocket hits the ground h=0, given that h=-16t²+320t+32
when h=0, our equation will be:
-16t²+320t+32=0
solving the above by completing square method we proceed as follows;
-16t²+320t+32=0
divide though by -16 we get
t²-20t-2=0
t²-20t=2
but
c=(-b/2a)^2
c=(20/2)^2
c=100
hence:
t²-20t+100=100+2
(t-10)(t-10)=102
√(t-10)²=√102
t-10=√102
hence
t=10+/-√102
t~20.1 or -0.1
since it must have taken long, then the answer is 20.1 sec
Answer:
(3 x + 2) (5 x - 4)
Step-by-step explanation:
Factor the following:
15 x^2 - 2 x - 8
Factor the quadratic 15 x^2 - 2 x - 8. The coefficient of x^2 is 15 and the constant term is -8. The product of 15 and -8 is -120. The factors of -120 which sum to -2 are 10 and -12. So 15 x^2 - 2 x - 8 = 15 x^2 - 12 x + 10 x - 8 = 5 x (3 x + 2) - 4 (3 x + 2):
5 x (3 x + 2) - 4 (3 x + 2)
Factor 3 x + 2 from 5 x (3 x + 2) - 4 (3 x + 2):
Answer: (3 x + 2) (5 x - 4)
