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2 years ago

Another way to express 52+24

1 answer:

4 0

We know that: 52 + 24 = 76.
You can also express it this way:
52 + 24 = 50 + 2 + 20 + 4 = ( 50 + 20 ) + ( 2 + 4 ) = 70 + 6 = 76
Or ( since 4 * 13 = 52 and 4 * 6 = 24 ):
52 + 24 = 4 * 13 + 4 * 6 = 4 * ( 13 + 6 ) = 4 * 19 = 76.

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Zoom im getting bored, Password: 833 7531 3785 Passcode: sC9BST

vodka [1.7K]

Answer:

how do we know if you are not a hacker

Step-by-step explanation:

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2 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

2 years ago

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$