<span>Let's say in general you want "n" adjacent squares. Each one needs a top, so that's n toothpicks. Each one needs a bottom, so that's another n toothpicks, and we're up to 2n total. You need one on the far left, and one on the far right, which brings us up to 2n + 2 toothpicks. Now we need to worry about the middle ones. But notice that you don't need n toothpicks for the interior because adjacent squares share. It turns out that you only need n-1. That brings us up to a grand total of 3n + 1 toothpicks, which would be 301 for n = 100.
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Answer:
oNE DAY A TRIANGLE SPLIT IN SO MANY DIFFERENT PEICES VBUT SUDDENLY ONE DAY THEY ALL CAME BACK.....
Step-by-step explanation:
Step-by-step explanation:
x+13÷2×2=4×2
x+13=8
x+13-13=8-13
x=8-13
x=-5
I've answered your other question as well.
Step-by-step explanation:
Since the identity is true whether the angle x is measured in degrees, radians, gradians (indeed, anything else you care to concoct), I’ll omit the ‘degrees’ sign.
Using the binomial theorem, (a+b)3=a3+3a2b+3ab2+b3
⇒a3+b3=(a+b)3−3a2b−3ab2=(a+b)3−3(a+b)ab
Substituting a=sin2(x) and b=cos2(x), we have:
sin6(x)+cos6(x)=(sin2(x)+cos2(x))3−3(sin2(x)+cos2(x))sin2(x)cos2(x)
Using the trigonometric identity cos2(x)+sin2(x)=1, your expression simplifies to:
sin6(x)+cos6(x)=1−3sin2(x)cos2(x)
From the double angle formula for the sine function, sin(2x)=2sin(x)cos(x)⇒sin(x)cos(x)=0.5sin(2x)
Meaning the expression can be rewritten as:
sin6(x)+cos6(x)=1−0.75sin2(2x)=1−34sin2(2x)
That is distributive property ... hope this helps
