Answer:1st one 2nd one and 8th
Step-by-step explanation:
The area of a rectangle is A=LW, the area of a square is A=S^2.
W=S-2 and L=2S-3
And we are told that the areas of each figure are the same.
S^2=LW, using L and W found above we have:
S^2=(2S-3)(S-2) perform indicated multiplication on right side
S^2=2S^2-4S-3S+6 combine like terms on right side
S^2=2S^2-7S+6 subtract S^2 from both sides
S^2-7S+6=0 factor:
S^2-S-6S+6=0
S(S-1)-6(S-1)=0
(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:
S=6 is the only valid value for S. Now we can find the dimensions of the rectangle...
W=S-2 and L=2S-3 given that S=6 in
W=4 in and L=9 in
So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.
Answer:
x=9
Step-by-step explanation:
Since x+5=14, then you would subtract 5 from 14. 14-5=9. So that means x equals 9.
Answer:
The second time when Luiza reaches a height of 1.2 m = 2 08 s
Step-by-step explanation:
Complete Question
Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.
H(t) = -0.6 cos (2pi/2.5)t + 1.5.
What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.
Solution
Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as
H(t) = -0.6cos(2π/2.5)t + 1.5
What is t when H = 1.2 m
1.2 = -0.6cos(2π/2.5)t + 1.5
0.6cos(2π/2.5)t = 1.2 - 1.5 = -0.3
Cos (2π/2.5)t = (0.3/0.6) = 0.5
Note that in radians,
Cos (π/3) = 0.5
This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,
Cos (5π/3) = 0.5
So,
Cos (2π/2.5)t = Cos (5π/3)
(2π/2.5)t = (5π/3)
(2/2.5) × t = (5/3)
t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.
Hope this Helps!!!
Answer:
answer to what? free points?
Step-by-step explanation:
