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2 years ago

Ya idk how to do this plz help explain if you can

Mathematics

1 answer:

Svet_ta [14]2 years ago

5 0

Answer:

92 attendees had activity cards

Step-by-step explanation:

Let x be the number of students with activity cards. Then 130-x is the number without, and the total revenue is ...

7x +10(130 -x) = 1024

7x +1300 -10x = 1024 . . . . eliminate parentheses

-3x = -276 . . . . . . . . . . . . . collect terms; subtract 1300

x = 92 . . . . . . divide by 3

92 students with activity cards attended the dance.

_____

<em>Comment on the solution</em>

Often, you will see such a problem solved using two equations. For example, they might be ...

Let 'a' represent the number with an activity card; 'w' the number without. Then ...

a+w = 130 . . . . the total number of students
7a +10w = 1024 . . . . the revenue from ticket sales

The problem statement asks for the value of 'a', so you want to eliminate w from these equations. You can do that using substitution. Using the first equation to write an expression for w, you have ...

w = 130-a

and making the substitution into the second equation gives ...

7a +10(130 -a) = 1024

This should look a lot like the equation we used above. There, we skipped the extra variable and went straight to the single equation we needed to solve.

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Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

k0ka [10]

I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

3 0

3 years ago

Plz help with math homework

Mkey [24]

Where is the question. though

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3 years ago

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Step-by-step explanation:

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3 years ago

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Answer:

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3 years ago

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cluponka [151]

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Answer:

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Step-by-step explanation:

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5 0

3 years ago