PLEASEEEE HELP??

Big Mack buys meat for the family reunion cookout. He needs to buy 11 packagesof meat. A package of hotdogs costs $2.59, and a p

IgorC [24]

Hello!

Let's write some important information contained in the exercise:

• hotdogs: $2.59 (,x,)

,

• hamburgers: $5.29 (,y,)

He needs 11 packages. Let's write it:

• x + y = 11

He spent a total of $39.29. We can write it as:

• 2.59x + 5.29y = 39.29

Now, let's solve these two equations as a linear system:

$\begin{cases}x+y=11\text{ equation A} \\ 2.59x+5.29y=39.29\text{ equation B}\end{cases}$

First, let's isolate x in equation A:

$\begin{gathered} x+y=11 \\ x=11-y \end{gathered}$

Now, we will replace where's x by 11-y in equation B:

$\begin{gathered} 2.59x+5.29y=39.29 \\ 2.59(11-y)+5.29y=39.29 \\ 28.49-2.59y+5.29y=39.29 \\ -2.59y+5.29y=39.29-28.49 \\ 2.7y=10.8 \\ y=\frac{10.8}{2.7} \\ \\ y=4 \end{gathered}$

As I called the hamburgers as 'y', we know that he bought 4 packages of hamburgers.

5 0

11 months ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$