I can model and solve one variable two step equations 9x-10= 63

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 $F(t)$ :

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

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

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

At these critical points, we get

$F(0)=1024$

$F\left(2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893$

$F\left(\dfrac\pi2\right)=4096$

$F(\pi)=-1024$

$F\left(\dfrac{3\pi}2\right)=4096$

$F\left(2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893$

We only care about 3 of these results.

$t=\dfrac\pi2\implies(x,y)=(0,8)$

$t=\pi\implies(x,y)=(-8,0)$

$t=\dfrac{3\pi}2\implies(x,y)=(0,-8)$

So to recap, we found that $f(x,y)$ attains

a maximum value of 4096 at the points (0, 8) and (0, -8), and
a minimum value of -1024 at the point (-8, 0).

6 0

3 years ago

Write the explicit formula for the following sequence. 3, 30, 300, 3000

masha68 [24]

Answer:

common ratio(r) = 10

place of the term = n

a = 3

now

sequence of 1 term = ar^(n-1)

= 3×10^(1-1)

= 3× 10^0

= 3×1

= 3

sequence of 2 term = ar^(n-1)

= 3×10¹

= 30

and so on

3 0

3 years ago

A 6 ounce package of fruit snacks contains 45 pieces. How many pieces would you expect in a 10 ounce package?

stiks02 [169]

First you must find the unit rate, so a 45 divided by 6 is 7.5, so then, multiply that by 10, 10 times 7.5 is 75, move the decimal place up, now you have the base rate, you can find any amount, so with these questions, all you must do is find the unit rate, in this case, the unit rate is 7.5, the unit rate will always be the ratio of one unit to the value of the unit, seen as ( 1-7.5 ) here.

Hope this helps, if not, comment below please!!!

7 0

4 years ago

Solve for c. C= [?] Enter the number that goes beneath the radical symbol.

photoshop1234 [79]

Answer:

c=45

Step-by-step explanation:

by the pythagorean theorem, c = sqrt(3^2+6^2) = sqrt(9+36) = sqrt(45)

Therefore, c=45.

5 0

3 years ago

Use Pythagorean Theorem to solve for the

babymother [125]

Pythagorean theorem is all about squaring the numbers. the equation is a^2+b^2=c^2
when you know two sides of the triangle, you fill in those spaces. they hypotenuse is ALWAYS ‘c’

to figure this problem out you put
a as 4
keep b as b since you don’t know what it is
c as 6

4 squared=16
6 squared=36

16+b^2=36
36-16=20
square root of 20 is 4.47

6 0

3 years ago