Answer:
2x to the second power +x−10
Step-by-step explanation:
Let's simplify step-by-step.
5x+x2−4−(4x−x2+6)
Distribute the Negative Sign:
=5x+x2−4+−1(4x−x2+6)
=5x+x2+−4+−1(4x)+−1(−x2)+(−1)(6)
=5x+x2+−4+−4x+x2+−6
Combine Like Terms:
=5x+x2+−4+−4x+x2+−6
=(x2+x2)+(5x+−4x)+(−4+−6)
=2x2+x+−10
ez copy pasta
Answer:
Step-by-step explanation:
The function for this problem is: h(t) = -16(t)^2 + vt + s h= the height t= time v= velocity s= starting height With the information given, we know that the starting height is 0, since it was from the ground, and the velocity of the ball is 35 feet per second. Inserting the these information into the equation, we get: h(t) = -16(t)^2 + 35t Now the question asks to find the maximum height. It can be done by using a grapher to graph the maximum of the parabola. It could also be done by finding the vertex, which would be the maximum, of the graph by using x= -b/(2a), where b is equal to 35 and a is equal to -16. We get x=35/32, the x-value of where the vertex lies. You can use this value as the t-value in the previous equation to find the h-value of the vertex. When you do, you get h= 19.1 feet, or answer D.
hope this helps
Answer:
hiduddseir rut the. rwi. ei ej ei rfcua di
12in/1foot and 3feet/yard
or 12in = 1foot
3feet= 2Yard
x intercept = (1,0)
y intercept = (0,-3)
if you have more problems you an use desmos.com and go to where it says f(x) expression
