Samantha wants to redecorate her room. She would like to replace the carpet and needs

Mathematics

2 answers:

CaHeK987 [17]2 years ago

6 0

There are No pictures

spayn [35]2 years ago

6 0

Lol you need pictures

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How many different ways can the letters of “football” be arranged

makvit [3.9K]

<h3>Answer: 10,080</h3>

Explanation:

There are 8 letters so there are 8! = 8*7*6*5*4*3*2*1 = 40,320 permutations of those letters. However, the letters "O" and "L" show up twice each, so we must divide by 2! = 2*1 = 2 for each instance this happens.

So,

(8!)/(2!*2!) = (40,320)/(2*2) = (40,320)/4 = 10,080

is the number of ways to arrange the letters of "football".

The reason we divide by 2 for each instance of a duplicate letter is because we can't tell the difference between the two "O"s or the two "L"s. If there was a way to distinguish between them, then we wouldnt have to divide by 2.

3 0

2 years ago

(1/4)^3z-1 =16^z+2*64^z-2 Z=___ Please help

wariber [46]

Answer:

Z = 0.198877274

Step-by-step explanation:

$(\frac{1}{4})^{3z-1} = 16^z + 2*16^{z-2}\\4^{1-3z} = 4^{2z} + 4^{\frac{1}{2}}*4^{2z-4}\\4^{1-3z} = 4^{2z} + 4^{2z-4+\frac{1}{2}}\\4^{1-3z} = 4^{2z} + 4^{2z-\frac{7}{2}}\\4^{1-3z} = 4^{2z} *(1+ 4^{-\frac{7}{2}})\\4^{1-3z} = 4^{2z} *(1+ 2^{-7})\\4^{1-3z} = 4^{2z} *(1+ \frac{1}{128} )\\4^{1-3z} = 4^{2z} *(\frac{129}{128} )\\Taking\;\; Logarithm\;\; with\;\; base\;\; 4\\Log_4(4^{1-3z}) = Log_4(4^{2z}) + Log_4(\frac{129}{128})\\1-3z = 2z + 0.005613627712 \\5z = 0.994386372\\z = 0.198877274$