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Tanya [424]
3 years ago
13

Hey! Can someone please help me with this question? Really appreciate it

Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

d = 0.112* 10^3

Step-by-step explanation:

Given

h = 8.4 * 10^3

d = \sqrt{\frac{3h}{2}}

Required

Find d

We have:

d = \sqrt{\frac{3h}{2}}

Substitute: h = 8.4 * 10^3

d = \sqrt{\frac{3*8.4 * 10^3}{2}}

d = \sqrt{\frac{25.2 * 10^3}{2}}

d = \sqrt{12.6 * 10^3}

Express as:

d = \sqrt{1.26 *10* 10^3}

d = \sqrt{1.26 *10^4}

Split

d = \sqrt{1.26} *\sqrt{10^4}

d = 1.122* 10^2

To write in form of: a * 10^b

The value of a must be: 0 \le a \le 1

So, we have:

d = 0.1122* 10 * 10^2

d = 0.1122* 10^3

Approximate

d = 0.112* 10^3

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A figure has an area of 64 unit square which describes the area of the figure after a dilation with a scale factor of 3/2
eduard

Answer:

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Step-by-step explanation:

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2 years ago
Find all the complex roots. Write the answer in exponential form.
dezoksy [38]

We have to calculate the fourth roots of this complex number:

z=9+9\sqrt[]{3}i

We start by writing this number in exponential form:

\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}

Then, the exponential form is:

z=18e^{\frac{\pi}{3}i}

The formula for the roots of a complex number can be written (in polar form) as:

z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}

<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>

Then, we calculate the arguments of the trigonometric functions:

\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})

We can now calculate for each value of k:

\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

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1 year ago
6 (3x-1)-10x. with work please :)
STatiana [176]
Hey there!
Let's break this expression into two parts:
6(3x-1) and -10x

To solve the first part, we need to use the distributive property which states:
a(b+c) = ab+ac

Applying that to this problem, we have:
6(3x) + 6(-1) =
18x - 6

Now, we can take that -10x and put it right back in:

18x - 6 - 10x

Combine like terms and subtract the 10x from the 18x to get:
8x - 6

Hope this helps!
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3 years ago
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lubasha [3.4K]
X^2-14x+49=0

Factor
(x-7)(x-7)

Check
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-7*-7=49

Use ZPP
x-7=0
x=7

Final answer: x=7
5 0
3 years ago
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