1.6022x 10-19 round three decimal places

On average, Shawnte drinks 1/2

jasenka [17]

Answer:

3/4 of a 6-ounce glass will be drunk by her in one hour.

Step-by-step explanation:

Given time = 2/3 of hour

= 2/3 (60) As hour = 60 mins

= 120/3

= 40 mins

Given quantity of water = 1/2 of 6-ounce

= 3 ounce

So,

In 40 mins, water intake is 3 ounce

In 1 min, water intake is 3/40

In 60 mins, water in take is (3/40)*60 = 4.5 ounce

Quantity of water intake in 60 mins:

4.5 ounce or 3/4 of 6-ounce glass

I hope it will help you.

3 0

3 years ago

Find the absolute maximum and absolute minimum of the function f(x,y)=2x3+y4f(x,y)=2x3+y4 on the region {(x,y)|x2+y2≤64}{(x,y)|x

ddd [48]

Check where the first-order partial derivatives vanish to find any critical points within the given region:

$f(x,y)=2x^3+y^4\implies\begin{cases}f_x=6x^2=0\\f_y=4y^3=0\end{cases}\implies(x,y)=(0,0)$

The Hessian for this function is

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}12x&0\\0&12y^2\end{bmatrix}$

with $\det\mathbf H(0,0)=0$ , so unfortunately the second partial derivative test fails. However, if we take $x=0$ we see that $f(x,y)>0$ for different values of $y$ ; if we take $y=0$ we see $f(x,y)$ takes on both positive and negative values. This indicates (0, 0) is neither the site of an extremum nor a saddle point.

Now check for points along the boundary. We can parameterize the boundary by

$(x,y)=(8\cos t,8\sin t)$

with $0\le t\le2\pi$ . This turns $f(x,y)$ into a univariate function $F(t)$ :

$F(t)=f(8\cos t,8\sin t)=2^{10}\cos^3t+2^{12}\sin^4t$

$\implies F'(t)=3\cdot2^{10}\cos^2t(-\sin t)+2^{14}\sin^3t\cos t=2^{10}\sin t\cos t(16\sin^2t-3\cos t)$

$F'(t)=0\implies\begin{matrix}\sin t=0\implies t=0,\,t=\pi\\\\\cos t=0\implies t=\dfrac\pi2,\,t=\dfrac{3\pi}2\\\\16\sin^2t-3\cos t=0\implies t=2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3},\,t=2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\end{matrix}$