Last two!!(ty for all the help) :

2 answers:

7 0

With the b2+b- 12 your gonna want to do this
b2+b-12
b2+4b-3b-12
which is the sum product
after doing that you wanna common the factors from the two pairs. Which is b2+4b-3b-12 then you do you parentheses around b(b+4)-3(b+4) just like that after you do that then you rewrite it in factored form b(b+4)-3(b+4) you then want to rewrite it to this (b-3)(b+4) after that your solution will be (b-3)(b+4) for the first one.

scoray [572]1 year ago

6 0

(B-3)(b+4) is the answer to the solution

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Write the square roots of 49 i in ascending order of the angle.

Troyanec [42]

Answer:

The square roots of 49·i in ascending order are;

1) -7·(cos(45°) + i·sin(45°))

2) 7·(cos(45°) + i·sin(45°))

Step-by-step explanation:

The square root of complex numbers 49·i is found as follows;

x + y·i = r·(cosθ + i·sinθ)

Where;

r = √(x² + y²)

θ = arctan(y/x)

Therefore;

49·i = 0 + 49·i

Therefore, we have;

r = √(0² + 49²) = 49

θ = arctan(49/0) → 90°

Therefore, we have;

49·i = 49·(cos(90°) + i·sin(90°)

By De Moivre's formula, we have;

$r \cdot (cos(\theta) + i \cdot sin(\theta) )^{\dfrac{1}{2}} = \pm \sqrt{r} \cdot \left (cos\left (\dfrac{\theta}{2} \right ) + i \cdot sin\left (\dfrac{\theta}{2} \right ) \right )$