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hammer [34]
1 year ago
13

Write the equation x^2 - 10x + 17 = 0 in the form (x+p)^2 = q (50 points!)

Mathematics
2 answers:
maxonik [38]1 year ago
7 0

Answer: (x-5)²=8

Step-by-step explanation:

x²-10x+17=0

x²-2(x)(5)+5²-5²+17=0

(x²-2(x)(5)+5²)-25+17=0

(x-5)²-8=0

(x-5)²-8+8=0+8

(x-5)²=8

snow_tiger [21]1 year ago
6 0

Answer:

(x - 5)² = 8

--------------------------

Given

Expression  x² - 10x +17.

Convert this to vertex form by completing the square.

Recall identity (a ± b)² = a² ± 2ab + b² and apply as given below:

x² - 10x + 17  =

x² - 2*5*x + 5² - 5² + 17  =

(x - 5)² - 25 + 17 = (x - 5)² - 8

(x - 5)² = 8

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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
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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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