What is the simplified form of the expression -4xy/3x -3y/4 -4x/3 -4y/3 -3x/4

If the dimensions of a rectangle are A plus 3b and A minus 3b what is the area of the rectangle

tresset_1 [31]

area=legnth times width

we notice this is a special case

(a+b)(a-b)=a^2-b^2

area=(a+3b)(a-3b)=a^2-9b^2

area=a^2-9b^2

5 0

2 years ago

Khan uses 78 feet of fence to make a rectangular pen for his pet rabbit. The length of the pen is 25 feet. What is the width of

Anna35 [415]

The width of the rectangular pen is equal to 14 feet.

7 0

2 years ago

0.2 ✕ -0.5 ✕ ( -0.9- -0.7)

Luba_88 [7]

Answer:

0.02

Step-by-step explanation:

$0.2\times -0.5\times (-0.9 - -0.7)= \\\\0.2\times -0.5 \times (-0.2)= \\\\0.02$

Hope this helps!

4 0

3 years ago

Read 2 more answers

Please help me asap, i was absent the day we learned this and haven't been able to pick it up

Maksim231197 [3]

Answer:

Your dead meat then...

Step-by-step explanation:

But I’m pretty sure its b?

sorry if it’s wrong

7 0

2 years ago

Read 2 more answers

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

3 years ago