4 years ago

Which values are solutions?

Mathematics

1 answer:

Mashutka [201]4 years ago

7 0

Answer:

D and E

Step-by-step explanation:

Given

$\sqrt{x}$ < 9 ( square both sides )

( $\sqrt{x}$ )² < 9², that is

x < 81

The only values where x < 81 are 75 and 80, that is D and E

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3+5i/-2+3i perform operation

melisa1 [442]

Answer:

$\dfrac{9}{13}-\dfrac{19}{13}i$

Step-by-step explanation:

Remember $i^2=-1$

You are given the fraction $\dfrac{3+5i}{-2+3i}$

First, multiply the numerator and the denominator by -2-3i:

$\dfrac{3+5i}{-2+3i}=\dfrac{(3+5i)(-2-3i)}{(-2+3i)(-2-3i)}=\dfrac{(3+5i)(-2-3i)}{(-2)^2-(3i)^2}=\dfrac{(3+5i)(-2-3i)}{4-9i^2}=\dfrac{(3+5i)(-2-3i)}{4+9}$

This gives you 13 in denominator, now multiply two complex numbers in numerator:

$(3+5i)(-2-3i)=-6-9i-10i-15i^2=-6-19i+15=9-19i$

Thus, the initial fraction is

$\dfrac{9-19i}{13}=\dfrac{9}{13}-\dfrac{19}{13}i$

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