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Elis [28]
3 years ago
11

Problem #1 -Find the equation of the line with the given slope and containing the given point.

Mathematics
1 answer:
DaniilM [7]3 years ago
3 0
For 1)

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ -7}}\quad ,&{{ 0}})\quad 
%   (c,d)

\end{array}
\\\quad \\\\ % slope  = m
slope = {{\boxed{ m}}}= \cfrac{rise}{run} \implies 
\implies -\cfrac{3}{5}
\\ \quad \\\\
% point-slope intercept
y-{{ y_1}}={{\boxed{ m}}}(x-{{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\
\qquad \uparrow\\
\textit{point-slope form}

for 2)

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ -3}}\quad ,&{{ 17}})\quad 
%   (c,d)
&({{ -7}}\quad ,&{{ 37}})
\end{array}
\\\quad \\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}
\\ \quad \\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\
\qquad \uparrow\\
\textit{point-slope form}

then change the y= part to f(x) = 

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tatiyna

Answer:

The solution to the system of equations is

\begin{gathered} x=\frac{179}{13} \\  \\ y=-\frac{279}{39} \\  \\ z=-\frac{48}{13} \end{gathered}

Explanation:

Giving the system of equations:

\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)

Subtracting (3) from 3 times (1), we have

3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

\begin{gathered} 13z=-48 \\  \\ z=-\frac{48}{13} \end{gathered}

Using the value of z obtained in (5), we have

\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\  \\ 3y+\frac{240}{13}=-3 \\  \\ 3y=-3-\frac{240}{13} \\  \\ 3y=-\frac{279}{13} \\  \\ y=-\frac{279}{39} \end{gathered}

Using the values obtained for y and z in (1), we have

\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\  \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\  \\ x-\frac{231}{13}=-4 \\  \\ x=-4+\frac{231}{13} \\  \\ x=\frac{179}{13} \end{gathered}

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1 year ago
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Answer:

For question 1:

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For question 2:

The expression for length a is

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Step-by-step explanation:

For question 1:

The length of hypotenuse expression =√(8^2+9^2)

For question 2:

Then expression for length a is

a=√(15^2-4^2)

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Step-by-step explanation:

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Where

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Answer:

The answer is 2 :)

Step-by-step explanation:

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