Emmy throws a doggy toy up in the air. The path of the toy is modeled by the function f(x)=−x^2+4x+5, where x is the number of feet the toy is from Emmy and f(x) is the height of the toy. What is the maximum height of the toy?

9feet

Step-by-step explanation:

Given the path of the toy modeled by the function f(x)=−x^2+4x+5, where x is the number of feet the toy is from Emmy and f(x) is the height of the toy.

AT maximum height, the velocity of the toy will be zero. Hence;

df(x)/dx = 0

-2x + 4 = 0

-2x = -4

x = -4/-2

x = 2

Get the maximum height;

Substitute x = 2 into the given function;

f(x)=−x^2+4x+5

f(2)=−2^2+4(2)+5

f(2) = -4+8+5

f(2) = 9feet

Hence the maximum height of the toy is 9feet

7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west