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2 years ago

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are any two regular polygons similar. The area of the smaller figure is about 390 cm^2. choose the best estimate of the area of

the larger figure

1 answer:

galben [10]2 years ago

4 0

Answer:

The best estimate of the area of the larger figure is $693\ cm^{2}$

Step-by-step explanation:

step 1

<em>Find the scale factor</em>

we know that

If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor

Let

z----> the scale factor

x-----> the corresponding side of the larger figure

y-----> the corresponding side of the smaller figure

so

$z=\frac{x}{y}$

we have

$x=12\ cm$

$y=9\ cm$

substitute

$z=\frac{12}{9}=\frac{4}{3}$ -----> the scale factor

step 2

<em>Find the area of the larger figure</em>

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z----> the scale factor

x-----> the area of the larger figure

y-----> the area of the smaller figure

so

$z^{2}=\frac{x}{y}$

we have

$z=\frac{4}{3}$

$y=390\ cm^{2}$

substitute and solve for x

$(\frac{4}{3})^{2}=\frac{x}{390}$

$(\frac{16}{9})=\frac{x}{390}$

$x=390*16/9=693.33\ cm^{2}$

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Use DeMoivre's Theorem to find the indicated power of the complex number. Write the answer in rectangular form.

slava [35]

Answer:

After solving the power:

$\bold{2(cos60^\circ+isin60^\circ)}$

Rectangular form:

$\bold{1+i\sqrt3}$

Step-by-step explanation:

Given the complex number:

$2(cos20^\circ+isin20^\circ)^3$

To find:

The indicated power by using De Moivre's theorem.

The complex number in rectangular form.

Rectangular form of a complex number is given as $a+ib$ where a and b are real numbers.

Solution:

First of all, let us have a look at the De Moivre's theorem:

$(cos\theta+isin\theta )^n=cos(n\theta)+isin(n\theta )$

First of all, let us solve:

$(cos20^\circ+isin20^\circ)^3$

Let us apply the De Moivre's Theorem:

Here, n = 3

$(cos20^\circ+isin20^\circ)^3 = cos(3 \times 20)^\circ+isin(3 \times 20)^\circ\\\Rightarrow cos60^\circ+isin60^\circ$

Now, the given complex number becomes:

$2(cos60^\circ+isin60^\circ)$

Let us put the values of $cos60^\circ = \frac{1}{2}$ and $sin60^\circ = \frac{\sqrt3}{2}$

$2(\dfrac{1}{2}+i\dfrac{\sqrt3}2)\\\Rightarrow (2 \times \dfrac{1}{2}+i\dfrac{\sqrt3}2\times 2)\\\Rightarrow \bold{1 +i\sqrt3 }$

So, the rectangular form of the given complex number is:

$\bold{1+i\sqrt3}$

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2 years ago

I need help with this geometry math assignment I been struggling and I need help on it very fast and it is due today so please h

photoshop1234 [79]

Answer:

12a. (-6, -6)

12b. 17

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13b. 2

Step-by-step explanation:

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$(\frac{-9-3}{2} ,\frac{2-14}{2} )$ = $(\frac{-12}{2} , \frac{-12}{2})$ = (-6, -6)

12b. Using the Distance formula:

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13a. x + 2y = 3

2y = 3 - x

y = $\frac{-1}{2}$ x + $\frac{3}{2}$

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2 years ago

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atroni [7]

Answer:

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Step-by-step explanation:

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2 years ago

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babunello [35]

The x-coordinate of the given endpoint will be " $x = 4 \pm 2\sqrt{29}$ ".

According to the question,

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Distance,

$6\sqrt{6}$

As we know,

The distance formula,

→ $Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

By substituting the values, we get

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→ $x-4=\pm 2\sqrt{29}$

→ $x = 4 \pm 2\sqrt{29}$

Thus the above answer is right.

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