Answer:
y > x - 2
y >= 5x + 1
Step-by-step explanation:
Answer:
24 area
18.80983669 perimeter
Step-by-step explanation:
1 (6*3/2)*2(1*3/2)+(4*3)
2 trust me
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
Answer:
12.0
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the trig function relating the adjacent side of an angle in a right triangle to the hypotenuse:
Cos = Adjacent/Hypotenuse
Filling in the given values, we have ...
cos(24°) = 11/x
Multiplying this equation by x/cos(24°), we get ...
x = 11/cos(24°) ≈ 12.0
Answer:
Yes
No
Yes
and
Yes
Step-by-step explanation:
Sorry if im wrong Happy Holidays
