Answer: I can’t solve (x+2)^2/64+(y-1)^2/81=1 but I solve (x+2)^2/64+(y-1)^2/81 it the same thing but I take out the =1
Step-by-step explanation: look at the picture and you see me answer, hope this help:)
Answer:
(n- 2/3)²
Step-by-step explanation:
- <em>Perfect square trinomial is: </em><em>a²+2ab+b²= (a+b)²</em>
We have:
It can be put as:
Here we consider n = a and -2/3 = b, then
Now we add 4/9 to a given binomial to make it perfect square:
- n² - 2×n×3/2 + 4/9= (n- 2/3)²
So, added 4/9 and got a perfect square (n- 2/3)²
Answer:
16 mls of the 75% solution and 20 mls of the 0.57% solution.
Step-by-step explanation:
Set up a system of equations:
Let x = volume of 75% solution and y be volume of 57% solution
0.75x + 0.57y = 36*0.65
0.75x + 0.57y = 23.4.......(1)
x + y = 36.........(2)
From equation (2):
y = 36 - x
Substituting this into equation(1):
0.75x + (0.57(36 - x) = 23.4
0.75x + 20.52 - 0.57x = 23.4
0.18x = 23.40 - 20.52 = 2.88
x = 2.88/0.18
x = 16
so from equation (2): y = 36-16 = 20.
Answer:
Step-by-step explanation:
Suppose at t = 0 the person is 1m above the ground and going up
Knowing that the wheel completes 1 revolution every 20s and 1 revolution = 2π rad in angle, we can calculate the angular speed
2π / 20 = 0.1π rad/s
The height above ground would be the sum of the vertical distance from the ground to the bottom of the wheel and the vertical distance from the bottom of the wheel to the person, which is the wheel radius subtracted by the vertical distance of the person to the center of the wheel.
(1)
where 
is vertical distance from the ground to the bottom of the wheel, 
is the vertical distance from the bottom of the wheel to the person, R = 10 is the wheel radius, 
is the vertical distance of the person to the center of the wheel.
So solve for in term of t, we just need to find the cosine of angle θ it has swept after time t and multiply it with R
Note that is negative when angle θ gets between π/2 (90 degrees) and 3π/2 (270 degrees) but that is expected since it would mean adding the vertical distance to the wheel radius.
Therefore, if we plug this into equation (1) then
