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3 years ago

Which pair of complex factors results in a real-number product?

Mathematics

2 answers:

Crank3 years ago

8 0

we will proceed to solve each case to determine the solution of the problem

we know that

$i^{2} =-1$

case A. $15*(-15i)$

$15*(-15i)=-225i$

The result of case A is not a real number product

case B. $3i*(1-3i)$

$3i*(1-3i)=3i*(1)-(3i)*(3i)\\=3i-9i^{2}\\=3i-9*(-1)\\=3i+9$

The result of case B is not a real number product

case C. $(8+20i)*(-8-20i)$

$(8+20i)*(-8-20i)=8*(-8)+8*(-20i)+20i*(-8)-20i*(20i)\\=-64-160i-160i-400i^{2}\\=-64-320i-400*(-1)\\=-64-320i+400\\=336-320i$

The result of case C is not a real number product

case D. $(4+7i)*(4-7i)$

$(4+7i)*(4-7i)=4^{2} -(7i)^{2}\\=16-49i^{2}\\=16-49*(-1)\\=65\\$

The result of case D is a real number product

therefore

the answer is the option

$(4+7i)*(4-7i)$

defon3 years ago

4 0

To answer the problem above, we must first define what "i" means. i means imaginary number which means square root of - 1, to get rid of i, you simply need to square, in this problem, we need to look for an equation where "i" is squared. The answer is D. (4+7i)(4-7i) to justify:
(4+7i)(4-7i)
(16-28i+28i-49i^2)
(16-49i^2)
16-49 = -33

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Question 2 of 5

AlekseyPX

Given:

The different recursive formulae.

To find:

The explicit formulae for the given recursive formulae.

Solution:

The recursive formula of an arithmetic sequence is $f(n)=f(n-1)+d, f(1)=a,n\geq 2$ and the explicit formula is $f(n)=a+(n-1)d$ , where a is the first term and d is the common difference.

The recursive formula of a geometric sequence is $f(n)=rf(n-1), f(1)=a,n\geq 2$ and the explicit formula is $f(n)=ar^{n-1}$ , where a is the first term and r is the common ratio.

The first recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+5$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 5. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(5)$

$f(n)=5+5(n-1)$

Therefore, the correct option is A, i.e., $f(n)=5+5(n-1)$ .

The second recursive formula is:

$f(1)=5$

$f(n)=3f(n-1)$ for $n\geq 2$ .

It is the recursive formula of a geometric sequence with first term 5 and common ratio 3. So, the explicit formula for this recursive formula is:

$f(n)=5(3)^{n-1}$

Therefore, the correct option is F, i.e., $f(n)=5(3)^{n-1}$ .

The third recursive formula is:

$f(1)=5$

$f(n)=f(n-1)+3$ for $n\geq 2$ .

It is the recursive formula of an arithmetic sequence with first term 5 and common difference 3. So, the explicit formula for this recursive formula is:

$f(n)=5+(n-1)(3)$