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4 years ago

Find all solutions of the equation. leave answers in trigonometric form x^3-8=0 2cis

Mathematics

1 answer:

Lera25 [3.4K]4 years ago

7 0

$x^3-8=0$

$x^3=8=8\mathrm{cis}\,0^\circ$

By DeMoivre's theorem, we have

$x=8^{1/3}\mathrm{cis}\left(\dfrac{0^\circ+360^\circ k}3\right)$

with $k=0,1,2$ . Then we have three solutions,

$x=2\mathrm{cis}\,0^\circ=2$

$x=2\mathrm{cis}\,120^\circ$

$x=2\mathrm{cis}\,240^\circ$

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2 years ago

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Rewrite the expression 2(x + 5) using the Distributive Property.

ehidna [41]

Answer:

2x + 10

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To expand (using the distributive property) multiply the number outside the bracket i.e. in this case '2', with the values inside the brackets.

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3 years ago

Can anyone help me find the (?,?) coordinates? The given shape is a parallelogram.

Sonja [21]

The missing coordinates of the parallelogram is (m + h, n).

Solution:

Given shape is a parallelogram.

Construction: Draw a line joining the diagonals.

In parallelogram, diagonals bisect each other.

Solve it using mid-point formula:

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Here $x_1=m, y_1=n, x_2=h, y_2=0$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n+0}{2}\right )$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n}{2}\right )$

Using this midpoint find the missing coordinate.

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Let the missing coordinates by x and y.

Here $x_1=0, y_1=0, x_2=x, y_2=y$

$$\text{Midpoint}=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{x}{2}, \frac{y}{2}\right )$

Now equate the x-coordinates and y-coordinates.

$$ \frac{m+h}{2}=\frac{x}{2}, \ \ \ \frac{n}{2} = \frac{y}{2}$

Multiply by 2 on both sides of the equation, we get

m + h = x, n = y

x = m + h and y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

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3 years ago