Why is it useful to identify or analyze occupational and industrial trends when considering your career path

2 answers:

8 0

When you are considering your professional future, sometimes we are undecided because we have a lot of options that seems perfect and matching with our skills and preferences. We have not decided yet what do we want to do for living.

Yet, the occupational and industrial trends are sometimes important information/aspects that we can review to make a decision if we are in troubles to decide. This trends contain information about employment, career exploration, education and training development plans and economic development. This can helps us to consider potential scenarios in our near future.

marin [14]3 years ago

5 0

It is useful. i mean because ull know how well ur going to do in the industry ur joining. if the industry is doing wee ur likely to be paid more.

You might be interested in

8_murik_8 [283]

It was a 15 percent tip

8 0

3 years ago

Read 2 more answers

Is (6, 2) a solution to this system of equations? x + 4y = 14 2x + 4y = 20

uranmaximum [27]

No it’s not because (6,2) would be a no solution so no

3 0

3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

7 0

3 years ago

What is the slope of a line perpendicular to the line with equation y = 1/6x – 2? help pls very urgent

valina [46]

Solution:

1) Simplify \frac{1}{6}x to \frac{x}{6}
y=\frac{x}{6}-2

2) Add 2 to both sides
y+2=\frac{x}{6}

3) Multiply both sides by 6
(y+2)\times 6=x

4) Regroup terms
6(y+2)=x

5) Switch sides
x=6(y+2)

Done!

7 0

3 years ago

Solve the quadratic equation by completing the square<br> x²+12x+26=0

vova2212 [387]

Answer:

x={-13,-2}

Step-by-step explanation:

(x+13)(x+2)=0

x+13=0

x=-13

x+2=0

x=-2

x={-13,-2}

4 0

3 years ago