i am thinking of a number.l take away 5. the result is 14 . what number did i think​

see the attachment hope this helps you

Answer:

See Below (Boxed Solutions).

Step-by-step explanation:

We are given the two complex numbers:

[tex]\displaystyle z = \sqrt{3} - i\text{ and } w = 6\left(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}\right)[/tex]

First, convert z to polar form. Recall that polar form of a complex number is:

[tex]z=r\left(\cos \theta + i\sin\theta\right)[/tex]

We will first find its modulus r, which is given by:

[tex]\displaystyle r = |z| = \sqrt{a^2+b^2}[/tex]

In this case, a = √3 and b = -1. Thus, the modulus is:

[tex]r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2[/tex]

Next, find the argument θ in [0, 2π). Recall that:

[tex]\displaystyle \tan \theta = \frac{b}{a}[/tex]

Therefore:

[tex]\displaystyle \theta = \arctan\frac{(-1)}{\sqrt{3}}[/tex]

Evaluate:

[tex]\displaystyle \theta = -\frac{\pi}{6}[/tex]

Since z must be in QIV, using reference angles, the argument will be:

[tex]\displaystyle \theta = \frac{11\pi}{6}[/tex]

Therefore, z in polar form is:

[tex]\displaystyle z=2\left(\cos \frac{11\pi}{6} + i \sin \frac{11\pi}{6}\right)[/tex]

Part A)

Recall that when multiplying two complex numbers z and w:

[tex]zw=r_1\cdot r_2 \left(\cos (\theta _1 + \theta _2) + i\sin(\theta_1 + \theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle zw = (2)(6)\left(\cos\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right) + i\sin\left(\frac{11\pi}{6} + \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{zw = 12\left(\cos\frac{9\pi}{4} + i\sin \frac{9\pi}{4}\right)}[/tex]

To find the complex form, evaluate:

[tex]\displaystyle zw = 12\cos \frac{9\pi}{4} + i\left(12\sin \frac{9\pi}{4}\right) =\boxed{ 6\sqrt{2} + 6i\sqrt{2}}[/tex]

Part B)

Recall that when raising a complex number to an exponent n:

[tex]\displaystyle z^n = r^n\left(\cos (n\cdot \theta) + i\sin (n\cdot \theta)\right)[/tex]

Therefore:

[tex]\displaystyle z^{10} = r^{10} \left(\cos (10\theta) + i\sin (10\theta)\right)[/tex]

Substitute:

[tex]\displaystyle z^{10} = (2)^{10} \left(\cos \left(10\left(\frac{11\pi}{6}\right)\right) + i\sin \left(10\left(\frac{11\pi}{6}\right)\right)\right)[/tex]

Simplify:

Simplify using coterminal angles. Thus, the polar form is:

[tex]\displaystyle \boxed{z^{10} = 1024\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)}[/tex]

And the complex form is:

[tex]\displaystyle z^{10} = 1024\cos \frac{\pi}{3} + i\left(1024\sin \frac{\pi}{3}\right) = \boxed{512+512i\sqrt{3}}[/tex]

Part C)

Recall that:

[tex]\displaystyle \frac{z}{w} = \frac{r_1}{r_2} \left(\cos (\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\right)[/tex]

Therefore:

[tex]\displaystyle \frac{z}{w} = \frac{(2)}{(6)}\left(\cos \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right) + i \sin \left(\frac{11\pi}{6} - \frac{5\pi}{12}\right)\right)[/tex]

Simplify. Hence, our polar form is:

[tex]\displaystyle\boxed{ \frac{z}{w} = \frac{1}{3} \left(\cos \frac{17\pi}{12} + i \sin \frac{17\pi}{12}\right)}[/tex]

And the complex form is:

[tex]\displaystyle \begin{aligned} \frac{z}{w} &= \frac{1}{3} \cos\frac{5\pi}{12} + i \left(\frac{1}{3} \sin \frac{5\pi}{12}\right)\right)\\ \\ &=\frac{1}{3}\left(\frac{\sqrt{2}-\sqrt{6}}{4}\right) + i\left(\frac{1}{3}\left(- \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right) \\ \\ &= \boxed{\frac{\sqrt{2} - \sqrt{6}}{12} -\frac{\sqrt{6}+\sqrt{2}}{12}i}\end{aligned}[/tex]

Part D)

Let a be a cube root of z. Then by definition:

[tex]\displaystyle a^3 = z = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

From the property in Part B, we know that:

[tex]\displaystyle a^3 = r^3\left(\cos (3\theta) + i\sin(3\theta)\right)[/tex]

Therefore:

[tex]\displaystyle r^3\left(\cos (3\theta) + i\sin (3\theta)\right) = 2\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right)[/tex]

If two complex numbers are equal, their modulus and arguments must be equivalent. Thus:

[tex]\displaystyle r^3 = 2\text{ and } 3\theta = \frac{11\pi}{6}[/tex]

The first equation can be easily solved:

[tex]r=\sqrt[3]{2}[/tex]

For the second equation, 3θ must equal 11π/6 and any other rotation. In other words:

[tex]\displaystyle 3\theta = \frac{11\pi}{6} + 2\pi n\text{ where } n\in \mathbb{Z}[/tex]

Solve for the argument:

[tex]\displaystyle \theta = \frac{11\pi}{18} + \frac{2n\pi}{3} \text{ where } n \in \mathbb{Z}[/tex]

There are three distinct solutions within [0, 2π):

[tex]\displaystyle \theta = \frac{11\pi}{18} , \frac{23\pi}{18}\text{ and } \frac{35\pi}{18}[/tex]

Hence, the three roots are:

[tex]\displaystyle a_1 = \sqrt[3]{2} \left(\cos\frac{11\pi}{18}+ \sin \frac{11\pi}{18}\right) \\ \\ \\ a_2 = \sqrt[3]{2} \left(\cos \frac{23\pi}{18} + i\sin\frac{23\pi}{18}\right) \\ \\ \\ a_3 = \sqrt[3]{2} \left(\cos \frac{35\pi}{18} + i\sin \frac{35\pi}{18}\right)[/tex]

Or, approximately:

[tex]\displaystyle\boxed{ a _ 1\approx -0.4309 + 1.1839i,} \\ \\ \boxed{a_2 \approx -0.8099-0.9652i,} \\ \\ \boxed{a_3\approx 1.2408-0.2188i}[/tex]

LOL

Answer:

D. The over 30s have a larger range and interquartile range than the under 30s

Step-by-step explanation:

In a data set, the range is the difference between the maximum and minimum. So, the range for under 30s is 20, while the range for over 30s is 24. Additionally, the interquartile range is the difference between Q3 and Q1. For a boxplot, Q3 is the line where the box ends and Q1 is the line where the box begins. Therefore, the IQR for the under 30s is 8, and the IQR for over 30s is 11. So, D must be correct.

Answer:

y = 10

Step-by-step explanation:

To find the y coordinate of the midpoint, take the y coordinates of the endpoints and average

(0+y)/2 = 5

Multiply each  die by 2

0+y = 10

y = 10

Answer:

1320

Step-by-step explanation:

Use the formula for sum of series, s(a) = n/2(2a + (n-1)d)

The terms increase by 6, so d is 6

a is the first term, -56

n is the terms you want to find, 33

Plug in the numbers, 33/2 (2(-56)+(32)6)

Simplify into 33(80)/2 and you get 1320

Answer:

So the sqrt(46) is between 6 and 7

Step-by-step explanation:

sqrt(46)

5*5 = 25

6*6=36

7*7=49

So the sqrt(46) is between 6 and 7

Basically count/add up the total amount of degrees that are include in the angle <FHD.

-- (central angles)

So, 35 + 65 = 100 degrees

We want to compute the probability P(A or B) because we could either solve it following path A, path B, or doing both paths.

P(A) = 1/3

P(B) = 2/5

P(A and B) = P(A)*P(B)  assuming A,B are independent

P(A and B) = (1/3)*(2/5)

P(A and B) = 2/15

--------

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 1/3 + 2/5 - 2/15

P(A or B) = 5/15 + 6/15 - 2/15

P(A or B) = (5 + 6 - 2)/15

P(A or B) = 9/15

We could reduce this to 3/5, but it appears your teacher has chosen not to.

Answer:

abc is a triangle so ,

a is ( 9,6 )

b is ( 9,3 )

and c is ( 3,3 )

Answer:

(-3)

Step-by-step explanation:

follow me if you want

Answer:

16

Step-by-step explanation:

4(3 - 1)^2

~Simplify using PEMDAS

4(2)^2

4(4)

16

Best of Luck!

The answer you are looking for is letter A, x=30.

Solution/Explanation:

First, write out the equation,

1200-5(3x+30)=600

Next, using the Distributive Property,

1200-15x-150=600

Simplify the left side of the equation just a little bit more,

1050-15x=600

Reverse order of terms on the left side, to make it a little bit easier to solve,

-15x+1050=600

Now, subtract 1050 from both sides,

-15x+1050-1050=600-1050

Now, simplify this part of the equation,

-15x=-450

Finally, divide both sides by -15,

So, therefore, the final answer is x=30.

I hope this helped you. Enjoy your day, and take care!

Answer:

? ≈ 37°

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos? = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{4}{5}[/tex] , then

? = [tex]cos^{-1}[/tex] ([tex]\frac{4}{5}[/tex] ) ≈ 37° ( to the nearest degree )

Answer:

range

Step-by-step explanation:

Answer:

B. range.

Step-by-step explanation:

others are:

» Standard variation.

» Interquatile range.

» Quatiles, deciles and percentiles.

» variance.

[tex]{ \underline{ \blue{ \sf{christ \: † \: alone}}}}[/tex]

Hi there!  

»»————-　★　————-««

I believe your answer is:  

[tex]\frac{1}{6}[/tex]

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Box Probability}}\\\\\rightarrow \frac{\text{# of boxed flawed}}{\text{# of boxes checked}} \\\\\rightarrow \frac{3}{18}\\\\\rightarrow \frac{3/3}{18/3}\\\\\rightarrow\boxed{\frac{1}{6}}[/tex]

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

252

Step-by-step explanation:

To be divisible by 3, it's digits have to add to a number that is a multiple of 3.

To be divisible by 4 its last 2 digits have to be divisible by 3.

So let's start with 1x1 which won't work because 1x1 is odd. so let's go to 2x2 and see what happens.

212 that's divisible by 4 but not 3

222 divisible by 3 but not 4

232 divisible by 4 but not 3

242 not divisible by either one.

252 I think this might be your answer

The digits add up to 9 which is a multiple of 3 and the last 2 digits are divisible by 4

Answer:

x=12

Step-by-step explanation:

LM = RS

9x+27 = 135

Subtract 27 from each side

9x+27-27 =135-27

9x=108

Divide each side by 9

9x/9 = 108/9

x = 12

the boxes are equivalent so the one with a single dash is equal to the other with a single dash.

the one with 2 dashes is equal to the other with 2 dashes so on and so forth

SR=LM

LM=9x+27

RS=135

9x+27=135

so I solve it in my own weird way but you can solve it differently. 135-27=108

108/9=12

so your answer is 12

Answer:

x(ax-1)-y(y+1)

Step-by-step explanation:

you have to group the like terms

ax^2-x-y^2-y

x(ax-1)-y(y+1)

I hope this helps

Answer:

Step-by-step explanation:

[tex]\frac{sinxcos^3x-cos xsin^3x}{cos^42x-sin^42x} \\=\frac{sin x cos x(cos^2x-sin ^2 x)}{(cis^2 2x+sin^2 2x)(cos^2 2x-sin ^22x)} \\=\frac{2sin x cos x cos 2x}{2(1)(cos 4x)} \\=\frac{sin 2x cos 2x}{2 cos 4x} \\=\frac{2 sin 2x cos 2x}{4 cos 4x} \\=\frac{sin 4x}{4 cos 4x} \\=\frac{1}{4} tan 4x[/tex]

Answer:

8n-2 = 6n+9

2n-2 = 9

2n = 11

n = 5.5

So C is correct

Let me know if this helps!

Answer:

20

Step-by-step explanation:

Since the triangles are similar then the sides must be proportional

10/6 = x/12 cross multiply expressions

6x = 120 divide both sides by 6

x = 20

Answer: x= 11√3= 19.0525 = 19.1

Step-by-step explanation:

Let the reference angle be 30

so

cos 30 = b/h

√3/2 = x/22

or, 22√3 = 2x

or. x = (22√3)/2

so, x = 11√3

Answer:

x = 19.1 cm

Step-by-step explanation:

→ Find the name of the side you are not given

Opposite

→ Find a formula without opposite in it

Cos = Adjacent ÷ Hypotenuse

→ Rearrange to make adjacent the subject

Adjacent = Cos × Hypotenuse

→ Substitute in the values

Adjacent = Cos ( 30 ) × 22

→ Simplify

Adjacent = 19.1

Answer:

Step-by-step explanation:

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

[tex]\frac{14}{30}[/tex] is not equivalent to [tex]\frac{2}{5}[/tex] in the list of fractions [tex]\frac{18}{45}, \frac{14}{30} , \frac{x}{y} \frac{10}{25}, \frac{8}{20}, \frac{16}{40}[/tex].

Equivalent fractions represent the same value even though they look different.

We know we can find an equivalent fraction of a given fraction by or dividing both the numerator and denominator of the given fraction with the same number (maybe LCM or HCF of the numerator or denominator).

[tex]\frac{18}{45}=\frac{18/9}{45/9}=\frac{2}{5}[/tex] (since [tex]9[/tex] is HCF of [tex]18, 45[/tex])

[tex]\frac{14}{30}=\frac{14/2}{30/2}=\frac{7}{15}[/tex] (since [tex]2[/tex] is HCF of [tex]7, 15[/tex])

[tex]\frac{10}{25} =\frac{10/5}{25/5} =\frac{2}{5}[/tex] (since [tex]5[/tex] is HCF of [tex]10, 25[/tex])

[tex]\frac{8}{20} =\frac{8/4}{20/4} =\frac{2}{5}[/tex] (since [tex]4[/tex] is HCF of [tex]8, 20[/tex])

[tex]\frac{16}{40} =\frac{16/8}{40/8} =\frac{2}{5}[/tex] (since [tex]8[/tex] is HCF of [tex]16, 40[/tex])

Thus, [tex]\frac{14}{30}[/tex] is not equivalent to [tex]\frac{2}{5}[/tex].

Learn more about Equivalent fractions here- https://brainly.com/question/17912

#SPJ2

The Constructed  confidence interval for the proportion of her photographs taken within the last year contained flowers is

[tex]CI\ E(0.57,0.65)[/tex]

From the question we are told that:

The proportion of photographs that had flowers P= 0.61

Margin of error of M.E= 0.04

Generally, the equation for Confidence interval for proportion is mathematically given by

[tex]CI=0.61 \pm 0.04[/tex]P \pm M.E

Therefore Confidence interval is

[tex]CI=0.61 \pm 0.04[/tex]

And can also be written as

[tex]CI\ E((0.61+0.04),0.61-0.04))[/tex]

[tex]CI\ E(0.57,0.65)[/tex]

In conclusion the Confidence interval is

[tex]CI\ E(0.57,0.65)[/tex]

For more information on this visit

https://brainly.com/question/24131141?referrer=searchResults

Answer:

hope it help you

Step-by-step explanation:

mark me brailiest answer

Answer:

149 inches squared

Step-by-step explanation:

top rectangle: 25 * 7 = 175

second rectangle: 8 * (25 - 17) = 8^2 = 64

triangle in bottom right: 1/2 * (13 - 8) * (15 - 11) = 10

175 + 64 + 10 = 149 sq in

hopefully got this right!

Answer:

216

Step-by-step explanation:

Volume of a rectangular prism = area of base * depth

Area of base: 72

Depth: 3

Volume = 72 * 3 = 216