HELP!! Jamaal bounces on a trampoline. His height, as a function of time, is modeled by y= -16x2(power of 2) + 20x + 4.

Mathematics

2 answers:

Harrizon [31]2 years ago

7 0

Answer:

B the function is nonlinear

Step-by-step explanation:

zhenek [66]2 years ago

3 0

Answer: B, the function is nonlinear

Step-by-step explanation:

You might be interested in

At the nature center, pond exploring began at 10:40 a.m. and lasted 2 hours and 28 minutes. Then lunch began and lasted 30 minut

Elan Coil [88]

It ended at
1:38 p.m.
Could you mark brainliest?

7 0

3 years ago

Read 2 more answers

Solve for x. Enter your answer in interval notation using grouping symbols <br> X^2 + 5x < 24

icang [17]

I think the correct answer would be : (-8,3）
I hope this helps !

7 0

3 years ago

Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .