Answer:
1.199 ≤ x ≤ 1.201
Step-by-step explanation:
You want ...
|y -7| ≤ 0.005
Substituting the expression for y, you have ...
|5x+1 -7| ≤ 0.005
|x -1.2| ≤ 0.001 . . . . . simplify, divide by 5
-0.0001 ≤ x -1.2 ≤ 0.001 . . . . . "unfold" the absolute value
1.199 ≤ x ≤ 1.201 . . . . . . . . . . . .add 1.2
Answer:
2-√x
Step-by-step explanation:
(4-x)/(2+√x)*(2-√x)/(2-√x)
(4-x)(2-√x)/4-x
=2-√x
Answer:
It takes 1 second for the tape to reach the ground.
Equation to use: ![y(t)=16-\frac{1}{2} g\,t^2[/tex with acceleration due to gravity "g" = 32ft/s^2]Step-by-step explanation:This is an object moving vertically under the action of the acceleration of gravity (32 ft/s^2), with a starting position of 16 feet, and with NO initial velocity (drops from the roof).The equation that describes the"y" position of the object as a function of time (t) will be written as:[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5B%2Ftex%20with%20acceleration%20due%20to%20gravity%20%22g%22%20%3D%2032ft%2Fs%5E2%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EStep-by-step%20explanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EThis%20is%20an%20object%20moving%20vertically%20under%20the%20action%20of%20the%20acceleration%20of%20gravity%20%2832%20ft%2Fs%5E2%29%2C%20with%20a%20starting%20position%20of%2016%20feet%2C%20and%20with%20NO%20initial%20velocity%20%28drops%20from%20the%20roof%29.%3C%2Fp%3E%3Cp%3EThe%20equation%20that%20describes%20the%22y%22%20%20position%20of%20the%20object%20as%20a%20function%20of%20time%20%28t%29%20will%20be%20written%20as%3A%3C%2Fp%3E%3Cp%3E%5Btex%5Dy%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
As explain above, the initial position
s 16 ft, there is no initial velocity, so
, and the acceleration of gravity is 32 ft/s^2, and should be considered negative [as pointing down in the y-direction], so the equation simplifies to:
![y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2\\y(t)=16-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5Cy%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
In order to find the tima it takes it to hit the ground, we simple solve the equation for t when y(t) = 0 (the tape has reached the ground (zero height in the y-direction):
![y(t)=16-\frac{1}{2} g\,t^2\\0=16-\frac{1}{2} g\,t^2\\-16=-\frac{1}{2} 32\,t^2\\-16=-16\,t^2\\t^2=1\\t=+/- 1 \,second](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5C0%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5C-16%3D-%5Cfrac%7B1%7D%7B2%7D%2032%5C%2Ct%5E2%5C%5C-16%3D-16%5C%2Ct%5E2%5C%5Ct%5E2%3D1%5C%5Ct%3D%2B%2F-%201%20%5C%2Csecond)
We select the positive time (+1 second) which is what makes physical sense, since a negative value in time would mean time before the tape was dropped.
So the answer is: It takes 1 second for the tape to reach the ground.
X=5
Do you need an explanation or is the answer enough?
Answer:
-8,-7,-6,-5
Step-by-step explanation:
to start pick 4 variables lets say x, y, z, and a then remember that consecutive means one after another so x = x ,y = x + 1, z = x + 2, and a = x + 3. x + y + z + a = -26 so x + x + 1 + x + 2 + x + 3 = 26 then solve for x. finally go back and solve for the other variables.