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Rashid [163]
3 years ago
13

Solve for x. 3:12 = x:16 9/4 4 64

Mathematics
2 answers:
NemiM [27]3 years ago
6 0

Answer:  The required value of x is 4.

Step-by-step explanation:  We are given to solve the following equation for the value of x :

3:12=x:16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

We know that

a:b=\dfrac{a}{b}.

So, from equation (i), we get

3:12=x:16\\\\\\\Rightarrow \dfrac{3}{12}=\dfrac{x}{16}\\\\\\\Rightarrow \dfrac{1}{4}=\dfrac{x}{16}\\\\\\\Rightarrow x=\dfrac{16}{4}\\\\\Rightarrow x=4.

Thus, the required value of x is 4.

Mekhanik [1.2K]3 years ago
3 0
You have
  3/12 = x/16
Multiply by 16
  16*(3/12) = x = 4

x = 4
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A box was 3/4 full of marbles. The box fell on the floor. Thirty marbles fell out. This was 20% of the marbles in the box. How m
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3 years ago
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
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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}}

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3 years ago
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