True if u have high or very low outlier in your data set it is generally preferred to use the median
The answer would be x^4 -16
Explanation: use the FOIL method.
(x^2 +4)(x^2 -4)= x^4 +4x -4x -16. 4x and -4x cancel each other out. You are left with x^4 -16.
F- irst
O- uter
I- nner
L- ast
36 cm^2
Step-by-step explanation:
<u>Small</u><u> </u><u>window</u>
Length: 2cm
Width: 2cm
<u>Area</u><u>:</u> 4 cm^2
<u>Big window</u>
Length: 4cm
Width: 3cm
<u>Area</u><u>:</u> 12 cm^2
Total area of the windows:
(Area of 4 small windows + area of 1 big window)
(4 cm^2 x 4 + 12cm^2)
= <u>28 cm^2</u>
<u>Above</u><u> </u><u>window</u><u> </u><u>(</u><u>approx</u><u>.</u><u>)</u>
<u>Rectangle</u>
Length: 3cm
Width: 2cm
<u>Area</u><u>:</u> 6 cm^2
<u>T</u><u>riangle</u>
Base: 1cm
Height: 1cm
<u>Area</u><u>:</u> 2 x 0.5 cm^2 = 1 cm^2
<u>Square</u><u> </u><u>(</u><u>between</u><u> </u><u>the</u><u> </u><u>triangles</u><u>)</u>
Length: 1cm
Width: 1cm
<u>Area</u><u>:</u> 1 cm^2
= 8 cm^2
<u>TOTAL</u><u> </u><u>AREA</u><u> </u><u>OF</u><u> </u><u>ALL</u><u> </u><u>WINDOWS</u>
= AREA OF 4 WINDOWS + AREA OF BIG WINDOW + AREA OF ABOVE WINDOW
= 16 cm^2 + 12 cm^2 + 8 cm^2
<h3>
= <u>
36 cm^2</u></h3>
<em>I</em><em> </em><em>hope</em><em> </em><em>I</em><em> </em><em>made</em><em> </em><em>the</em><em> </em><em>explanations</em><em> </em><em>clear</em><em> </em><em>enough</em><em> </em><em>to</em><em> </em><em>make</em><em> </em><em>it</em><em> </em><em>easier</em><em> </em><em>for</em><em> </em><em>you</em><em> </em><em>to</em><em> </em><em>understand</em><em>!</em>
