The 400th term is 425.There are floor(√400) = 20 squares in the range 1..400, so the 400th term will be at least 420. There are floor(∛420) = 7 cubes in the range 1..400, so the 400th term may be as high as 427. However, there are
![\lfloor\sqrt[6]{427}\rfloor=2](https://tex.z-dn.net/?f=%5Clfloor%5Csqrt%5B6%5D%7B427%7D%5Crfloor%3D2)
numbers that are both squares and cubes. Consequently, the 400th term will be 427-2 =
425.
Answer:
42.5
Step-by-step explanation:
Multiply 50 by 0.85, which equals 42.5.
Answer: 12 = r
Step-by-step explanation: When we have this kind of a setup, we want to put our variables together on one side of the equation and our numbers together on the other side of the equation.
First, let's put our variables on the right side by subtracting 3r from both sides of the equation. That gives us 5 = 2r - 19.
Now we can move our numbers to the left by adding 19 to both sides of the equation and we get 24 = 2r.
Divide both sides by 2 and 12 = r
Note:
Don't just do this problem in your head. It's extremely important to develop the habit of putting all your steps down on paper or digitally. It will really pay off for you down the line.
Answer:
H = V0y t - 1/2 g t^2 equation for vertical height of object with initial speed (V0y = V0 sin theta)
If H is to be considered an absolute value from t = 0
h = H + 3 = V0y t - 1/2 g t^2 + 3 where h is height from ground
