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3 years ago

14

Please help must show workthere are 5 that I'm stuck on

1 answer:

Cloud [144]3 years ago

7 0

you cannot show too much "work"

basically, you remove what is common to all of the factors, and then put brackets, as it will be multiplied back in, remember that when you multiply exponents with the same base, its same as adding them, so subtract to remove...

you can seperate two of the variables , then factor, then subtract the last one from those two, because it cannot be factored out , as in part2 #2

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What is the value of i 97 – i?<br> –i<br> 0<br> –2i<br> i Superscript 96

Valentin [98]

<em>The question is not clearly readable, but I'm assuming the expression which makes more sense, so you can have a clue</em>

Answer:

$i^{97}-i=0$

Step-by-step explanation:

<u>Powers of The Imaginary Unit</u>

The imaginary unit i is defined as

$i=\sqrt{-1}$

The first powers of i are

$i^0=1$

$i^1=i$

$i^2=(\sqrt{-1})^2=-1$

$i^3=i.i^2=-i$

$i^4=i^2.i^2=1$

And so on the cycle repeats every four numbers. To find the value of $i^{96}$ we can find the remainder of 96/4=0. So $i^{96}=i^0=1$

The given expression is

$i^{97}-i$

Factoring

$i(i^{96}-1)$

Since $i^{96}=1$

$=i(1-1)=0$

The required value is 0

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The answer to this is -9

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3 years ago

Choose all the functions that give the sequence:

TEA [102]

Hence, the functions that produce given sequence are:

f(n) = 2n + 1

f(n) = 2(n − 1) + 3

Further explanation:

We will put n=1,2,3,4,5 to find which functions give the given sequence

<u>f(n) = 2n − 1</u>

Putting values of n

$f(1)=2(1)-1=2-1=1\\f(2)=2(2)-1=4-1=3\\f(3)=2(3)-1=6-1=5$

This function doesn't generate the given sequence

<u>f(n) = 2n + 1</u>

Putting values of n

$f(1)=2(1)+1=2+1=3\\f(2)=2(2)+1=4+1=5\\f(3)=2(3)+1=6+1=7\\f(4)=2(4)+1=8+1=9\\f(5)=2(5)+1=10+1=11$

This function generates the given sequence.

<u>f(n) = 2(n − 1) − 3</u>

Putting values of n

$f(1) = 2(1 -1) -3=2(0)-3=0-3=-3\\f(2) = 2(2 -1) -3=2(1)-3=2-3=-1\\f(3) = 2(3-1) -3=2(2)-3=4-3=1$

This function doesn't generate the given sequence

<u>f(n) = 2(n − 1) + 3</u>

Putting values of n

$f(1) = 2(1 - 1) + 3=2(0)+3=0+3=3\\f(2) = 2(2 - 1) + 3=2(1)+3=2+3=5\\f(3) = 2(3 - 1) + 3=2(2)+3=4+3=7\\f(4) = 2(4 - 1) + 3=2(3)+3=6+3=9$

Hence, the functions that produce given sequence are:

f(n) = 2n + 1

f(n) = 2(n − 1) + 3

Keywords: Functions, Sequence

Learn more about functions at:

#LearnwithBrainly

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3 years ago