The easiest way would be adding 76
The maximum height of the projectile is the maximum point that can be gotten from the projectile equation
The projectile reaches the maximum height after 5 seconds
The function is given as:
Differentiate the function with respect to t
Set to 0
So, we have:
Collect like terms
Solve for t
Hence, the projectile reaches the maximum after 5 seconds
Read more about maximum values at:
brainly.com/question/6636648
Answer:
The magnitude is 
The direction is 
i.e toward the x-axis
Step-by-step explanation:
From the question we are told that
The function is 
The point considered is 
Generally the maximum rate of change of f at the given point and the direction is mathematically represented as
![\Delta f(x,y) = [\frac{\delta f(x,y)}{\delta x } i + \frac{\delta f(x,y)}{\delta y } j ]](https://tex.z-dn.net/?f=%5CDelta%20f%28x%2Cy%29%20%3D%20%20%5B%5Cfrac%7B%5Cdelta%20%20f%28x%2Cy%29%7D%7B%5Cdelta%20x%20%7D%20i%20%2B%20%5Cfrac%7B%5Cdelta%20%20f%28x%2Cy%29%7D%7B%5Cdelta%20y%20%7D%20j%20%20%5D)
![\Delta f(x,y) = [\frac{\delta (9sin(xy))}{\delta x} i + \frac{\delta (9sin(xy))}{\delta y} i ]](https://tex.z-dn.net/?f=%5CDelta%20f%28x%2Cy%29%20%3D%20%20%5B%5Cfrac%7B%5Cdelta%20%20%289sin%28xy%29%29%7D%7B%5Cdelta%20x%7D%20i%20%20%2B%20%5Cfrac%7B%5Cdelta%20%20%289sin%28xy%29%29%7D%7B%5Cdelta%20y%7D%20i%20%20%20%5D)
![\Delta f(x,y) = [9y cos (x,y) i + 9xcos (x,y) j]](https://tex.z-dn.net/?f=%5CDelta%20f%28x%2Cy%29%20%3D%20%5B9y%20cos%20%28x%2Cy%29%20i%20%2B%20%209xcos%20%28x%2Cy%29%20j%5D)
At
![\Delta f (0,8) = [9(8) cos(0* 8)i + 9(8) sin(0* 8)j ]](https://tex.z-dn.net/?f=%5CDelta%20%20f%20%280%2C8%29%20%3D%20%20%5B9%288%29%20cos%280%2A%208%29i%20%20%2B%209%288%29%20sin%280%2A%208%29j%20%20%5D)
Answer: 145
Step-by-step explanation: I say this because it has 14. 4 then
Answer:
The center is -1,5 and the radius is 2
Step-by-step explanation:
Subtract 22 from both sides of the equation. x 2 + y 2 + 2 x − 10 y = − 22 Complete the square for x 2 + 2 x . ( x + 1 ) 2 − 1 Substitute ( x + 1 ) 2 − 1 for x 2 + 2 x in the equation x 2 + y 2 + 2 x − 10 y = − 22 . ( x + 1 ) 2 − 1 + y 2 − 10 y = − 22 Move − 1 to the right side of the equation by adding 1 to both sides. ( x + 1 ) 2 + y 2 − 10 y = − 22 + 1 Complete the square for y 2 − 10 y . ( y − 5 ) 2 − 25 Substitute ( y − 5 ) 2 − 25 for y 2 − 10 y in the equation x 2 + y 2 + 2 x − 10 y = − 22 . ( x + 1 ) 2 + ( y − 5 ) 2 − 25 = − 22 + 1 Move − 25 to the right side of the equation by adding 25 to both sides. ( x + 1 ) 2 + ( y − 5 ) 2 = − 22 + 1 + 25 Simplify − 22 + 1 + 25 . ( x + 1 ) 2 + ( y − 5 ) 2 = 4 This is the form of a circle. Use this form to determine the center and radius of the circle. ( x − h ) 2 + ( y − k ) 2 = r 2 Match the values in this circle to those of the standard form. The variable r represents the radius of the circle, h represents the x-offset from the origin, and k represents the y-offset from origin. r = 2 h = − 1 k = 5 The center of the circle is found at ( h , k ) . Center: ( − 1 , 5 ) These values represent the important values for graphing and analyzing a circle.
