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3 years ago

What is the product of complex numbers (-3i +4) and (3i +4)??? please explain

2 answers:

7 0

Product of complex numbers (-3i +4) and (3i +4)

(x - y)(x + y) = x² - y²

(-3i +4)(3i +4) = (4 -3i )(4 + 3i) = 4² - (3i)² = 16 - 9*(-1)=16 + 9 = 25

drek231 [11]3 years ago

6 0

Answer:

$(-3i+4)*(3i+4)=25$

Step-by-step explanation:

1. First writhe the complex numbers to multiply:

$(-3i+4)*(3i+4)$

2. Multiply the first term of the first parentheses by the first term of the second parentheses:

$-3i*3i=-9i^{2}$

3. Multiply the first term of the first parentheses by the second term of the second parentheses:

$-3i*4=-12i$

4. Multiply the second term of the first parentheses by the first term of the second parentheses:

$4*3i=12i$

5. Multiply the second term of the first parentheses by the second term of the second parentheses:

4*4=16

6. Write all the terms:

$(-3i+4)*(3i+4)=-9i^{2}-12i+12i+16$

7. Add up the terms:

$(-3i+4)*(3i+4)=-9i^{2}+16$

8. As $i^{2}=-1$ :

$(-3i+4)*(3i+4)=-9(-1)+16\\(-3i+4)*(3i+4)=9+16\\(-3i+4)*(3i+4)=25$

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3 years ago

A cheerleading squad consists of ten cheerleaders of ten different heights. How many ways are there for the cheerleaders to line

Sindrei [870]

Answer:

11340

Step-by-step explanation:

We have 10 cheerleaders of ten different heights. There are two rows each having 5 positions so that five cheerleaders occupy the first row that have height less than the five cheerleaders who occupied the second row.

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