Which of the following is equivalent to the complex number i24?

Mathematics

1 answer:

Alika [10]3 years ago

5 0

Answer: 1

Step-by-step explanation:

Rewrite as (i^4)^6

Rewrite i^4 as 1

Rewrite i^4 as (i^2) ^2

((i^2)) ^2)6

Rewrite i^2 as -1

((-1) ^2) 6

Raise -1 to the power of 2

1^6

One to any power is 1.

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The formula for the area of a triangle is A = one-half b h. Which is the equivalent equation solved for h?

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3 years ago

How to evaluate question number 4

ANTONII [103]

There are 2 ways that spring to mind.

One is to use the definition of the derivative at a point:

$f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

In this case, $c=0$ and $f(x)=\sqrt{5-x}$ . Then

$f'(x)=-\dfrac1{2\sqrt{5-x}}\implies f'(0)=\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\dfrac1{2\sqrt5}$

- - -

If you don't know about derivative yet (I would think you do, considering this is for a midterm in AP Calc, but I digress), the other way is to rely on algebraic manipulation. Multiply the numerator and denominator by the conjugate of the numerator to get

$\dfrac{\sqrt{5-x}-\sqrt5}x\cdot\dfrac{\sqrt{5-x}+\sqrt5}{\sqrt{5-x}+\sqrt5}=\dfrac{\left(\sqrt{5-x}\right)^2-\left(\sqrt5\right)^2}{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac x{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac1{\sqrt{5-x}+\sqrt5}$

This is continuous at $x=0$ , so the limit is the value of the expression at $x=0$ :

$\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\lim_{x\to0}\frac1{\sqrt{5-x}+\sqrt5}=-\frac1{2\sqrt5}$

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3 years ago

Answer the question from the picture below

labwork [276]

We can see that the slope of this line is -2 and that there is no y intercept. The inequality for this line should be $y \geq -2$ . This is because the shaded side is on the right side of the line, indicating all numbers greater than or equal to the line that satisfy the inequality. If it were a dotted line then it would be $y > -2$ , but since it isn't, your answer is $y \geq -2$ . Hope this helps.