Which shows one way the equation can be represented in words? Z-6=1.4

Find the solution of Z^5=32 whose argument is strictly between 0 and 90 degrees

tigry1 [53]

Answer:

$\|r\| = 32$ , $\theta = 0\,rad$

Step-by-step explanation:

This complex expression is computed from the De Moivre expression:

$z^{5} = 32\cdot (\cos 5x + i \cdot \sin 5x)$

Given that resultant expression is entirely real, the expression is simplified into this form:

$32 = 32\cdot \cos 5x$

$\cos 5x = 1$

The angle is 0 radians.

8 0

3 years ago

(Algebra 2 Conics):

olchik [2.2K]

$\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ 4p(y- k)=(x- h)^2 \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ -\cfrac{1}{4}(y+2)^2=x-7\implies -\cfrac{1}{4}[y-(-2)]^2=x-7 \\[2em] [y-(-2)]^2=-4(x-7)\implies [y-(\stackrel{k}{-2})]^2=4(\stackrel{p}{-1})(x-\stackrel{h}{7})$

so h = 7, k = -2, meaning the vertex is at (7, -2).

the squared variable is the "y", meaning is a horizontal parabola.

the "p" distance is negative, for a horizontal parabola that means, it's opening towards the left-hand-side.

we know the focus and directrix are "p" units away from the vertex, and we know the parabola is opening horizontally towards the left-hand-side.

the focus is towards it opens 1 unit away, at (6, -2).

the directrix is on the opposite direction, 1 unit away, at (8, -2), namely x = 8.

8 0

3 years ago

Question set 1: One instructor believes that students take more than 2 classes per quarter on average. He randomly interviewed a

Sveta_85 [38]

Answer:

There is statistical evidence at 95% level to claim that students attend more than 2 classes per quarter

Step-by-step explanation:

Given that one instructor believes that students take more than 2 classes per quarter on average. Let X be the no of classes students take.

$H_0: \bar x =2\\H_a: \bar x >2$

(Right tailed test at 5% significance level)

$\bar x =2,3 \\s = 0.8$

Mean difference = $2.3-2=0.3$

Std error = $\frac{0.8}{\sqrt{24} } \\=0.1633$

Test statistic t = mean diff/std error = 1.84

p value =0.039

Since p <0.05, we reject null hypothesis.

There is statistical evidence at 95% level to claim that students attend more than 2 classes per quarter

3 0

4 years ago

1 × 6,391 _____ 0.1 × 6,391 ____ 0.01 × 6,391 _____

Cerrena [4.2K]

1.= 6,391

2.=6,3910

3. = 6,39100

All you do is add the 0s at the end

5 0

3 years ago

Verify that the following equation is an identity. (1/sinx) - (1/cscx) = cscx - sinx

Anarel [89]

Manipulating the left side of the equation, we obtain:

$\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}$

Using that $\csc x=\dfrac{1}{\sin x}$ :

$\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\dfrac{1}{\sin x}}=\dfrac{\csc x-\sin x}{1}=\csc x-\sin x\\\\\boxed{\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\csc x-\sin x}~~\blacksquare$

6 0

3 years ago

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