How does poetry play an important role in mathematics?

1 answer:

8 0

Poetry has always been used as a vehicle of enlightenment, By expressing mathematics formula poetically, you can impact knowledge across the board.<span />

You might be interested in

7. Expand and reduce A(x). Then calculate A(x) for x=1.<br> A (x) = (3x – 3)(3x + 2)

SashulF [63]

Given:

The function is

$A(x)=(3x-3)(3x+2)$

To find:

The simplified form of A(x) and value of A(x) at x=1.

Solution:

We have,

$A(x)=(3x-3)(3x+2)$

$A(x)=(3x)(3x)+(3x)(2)+(-3)(3x)+(-3)(2)$

$A(x)=9x^2+6x-9x-6$

$A(x)=9x^2-3x-6$

Putting x=1, we get

$A(1)=9(1)^2-3(1)-6$

$A(1)=9-3-6$

$A(1)=9-9$

$A(1)=0$

Therefore, the simplified form of A(x) is $A(x)=9x^2-3x-6$ and the value of A(x) at x=1 in 0.

4 0

3 years ago

The measures of the angles of a triangle are shown in the figure below. Solve for x.

-Dominant- [34]

Answer:

33 degrees

Step-by-step explanation:

total degrees must add up to 180

106+41=147

180-147=33 degrees

6 0

3 years ago

Write an equation of a line that is perpendicular to the line y=2/3x and passes through origin

sesenic [268]

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?

$\bf y = \cfrac{2}{3}x\implies y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+0\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}$

so we're really looking for the equation of a line whose slope is -3/2 and runs through (0,0).

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{3}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{3}{2}x$