The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet.
We know there will be a constant downward acceleration of 32.15 ft/s^2.
Use the following equation:
v_final^2 = v_initial^2 + 2ah
v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2
v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.
2876.74665 in
After the in you got to put a number 2 above it I don't know how to make the symbol.
There are 5 solutions for this system.
x^2 + 4y^2 = 100 ____1
4y - x^2 = -20 ____2
Add both 1 & 2 together. x^2 gets cancelled
4y^2 + 4y = 80 (send 80 to the other side and divide by 4)
Then equation the becomes : y^2 + y -20 =0
Now factorise the equation: (y+5) (y-4) = 0
Solve for y : y = -5 and y = 4
Using the values of y to find the values of x. From equation 1:
x^2 = 100 - 4y^2 x = /100 - 4y^2 (/ means square root) Replace values of y
y = -5, x = /100 - 4(-5)^2 = /100 - 100 = 0
y = 4, x = /100 - 4(4)^2 = / 100 - 64 = /36 = -6 or 6
Thus we have 6 solutions y = -5, 4 and x = -6, 0, 6
Answer:
18
Step-by-step explanation:
1/3 * 18 = 6 or 6/(1/3) = 18 or 6 * 3 =18
