<span>(5+2 i)(4-3i) - (5-2yi)(4-3i)
Factorize out (4 -3i)
(4 -3i)( (5 +2i) - (5 -2yi) )
= </span><span><span>(4 -3i)(5 +2i - 5 + 2yi)</span>
= </span><span><span>(4 -3i)(5 - 5 + 2i + 2yi)</span>
= (4 -3i)(2i + 2yi)
= (4 - 3i)(2 + 2y)i. Let's multiply the first two.
</span>
(4 - 3i)(2 + 2y) = 2*(4 -3i) + 2y*(4 - 3i)
= 8 - 6i + 8y - 6yi
= 8 + 8y - 6i - 6yi
(4 - 3i)(2 + 2y)i = (8 + 8y - 6i - 6yi)i Note: i² = -1
= 8i + 8yi - 6i² - 6yi²
= 8i + 8yi - 6(-1) - 6y(-1)
= 8i + 8yi + 6 + 6y
= 6 + 6y + 8i + 8yi
= (6 + 6y) + (8 + 8y)i In the form a + bi
Answer:
10 more girls
Step-by-step explanation:
So here 2/3 of 24 ran for first race
Then it gives us that 16 girls ran for first race.
Next, 1/4 of 24 ran for second race
That is 6 girls
The remaining girls ran for the third race,
24 - 22
2 girls
16 - 6 = 10
So for first race 10 more girls ran
Start by combining the fractions:
Recall the Pythagorean identity:
Then cancel a factor of 
and use the definition of secant:
Answer:
B
Step-by-step explanation:
Use the distance formula in 3 dimensions
d = 
with (x₁, y₁, z₁ ) = (3, 4, 10) and (x₂, y₂, z₂ ) = (8, 4, - 2)
d = 
= 
= 
= 
= 13 → B
