What are the real and imaginary parts of z? Z = 20i+26

Answer: 52x+4

Step-by-step explanation: Do 16(2x-1)+4(5x+5) and get 52x+4.

Answer:

6 6 6 6 11

Step-by-step explanation:

The median is the middle number

the mode is the number of items, usually the number that are the same.

the mean is the average

the range is the number between the smallest and the largest.

So the information you have makes 4 of the 5 numbers the same.

The middle number is 6. This is an interesting fact because it means that 4 of the five numbers are 6

x 6 6 6 6  or  6 6 6 6 x  is what you know so far.

The range is a bit of a devil. And so is the mean. We need the mean to be related to  5*7 = 35. So the sum of the five numbers equals 35. The range would have to be 5, I think. (35 - 4*6) = 11

The difference between 11 and 6 = 5

So 11 works for everything but the range.

If the range must be ten, then x = 16 but then the mean = 8 and 7  is incorrect

So I can get 3 out of 4, but not them all.

Answer:

28 km per hour

Step-by-step explanation:

First determine the amount of time that the trip took

17:55 - 17:40 = :15 or 15 minutes

We need the time in hours

15 minutes * 1 hour/ 60 minutes = 15/60 hours = 1/4 hours

To determine speed, take distance and divide by time

7 km ÷( 1/4 hours)

Copy dot flip

7 * 4/1

28 km per hour

Answer:

complete the square 2 x + 38 = 0

Step-by-step explanation:

Answer:

.............soooooooooooooooerrrrrrrrrrrrryyyyyyyyyyyyyyy.......... iiiiiiiii........ doooooooonnnnnnnnntttttttt......... kkkknnnnnooooowwww

Answer:

to do this lets first solve this so

5*5*5*5=625

5^-3=0.008

0.008^2=0.000064

0.000064*625=0.4

so all equations that are equal to 0.4 are right which is C

Answer:

The answer is below

Step-by-step explanation:

The area of the wall is 84 ft².

The width of the wallpaper is 18 in.

1 ft. = 12 in

Hence; 18 in = 18 in * 1 ft. per 12 in = 1.5 ft

The length of the wallpaper is 33 ft.

Therefore, the area of the wallpaper = length * width = 1.5 ft. * 33 ft. = 49.5 ft². This means that each roll of wallpaper has an area of 49.5 ft²

Therefore, the minimum amount of rolls of wallpaper needed = area of wall / area of wallpaper

Amount of wallpaper = 84 ft² / 49.5 ft² = 1.7 rolls

Answer:

A: 42

Step-by-step explanation: 10.5x4=42

Answer:

your answer is A 42

Step-by-step explanation:

make me brainliest

Answer:

1

Step-by-step explanation:

that's a very vague question

Answer:

what do you mean for this question. What do you need help on?

The lower the frequency the least common the average.

There are two frequencies that are 1, which would be the least common.

The answer is C. 0.320-0.329 and 0.360 -0.369

Answer:

option C

Step-by-step explanation:

option c is contains the two batting averages which have a frequency of 1, this means that they both only occurred once which is the least amount of times any of the batting averages has occurred.

Answer:

See Explanation

Step-by-step explanation:

You posted an incomplete question with little and unclear details.

I will answer this question with a more complete version (see attachment)

Given that:

[tex]Scores: 100\ 95\ 88\ 62\ 76\ 90\ 100\ 58\ 72\ 60\ 85\ 90\ 70\ 72\ 54\ 100\ 60\ 80\ 75\ 51[/tex]

Required

The fraction that passed the test

Rearrange the score in ascending order:

[tex]Scores: 51\ 54\ 58\ 60\ 60\ 62\ 70\ 72\ 72\ 75\ 76\ 80\ 85\ 88\ 90\ 90\ 95\ 100\ 100\ 100[/tex]

The total number of students is:

[tex]n =20[/tex]

Extract the students that passed (scored 70 and above)

[tex]Passed: 70\ 72\ 72\ 75\ 76\ 80\ 85\ 88\ 90\ 90\ 95\ 100\ 100\ 100[/tex]

Their numbers are:

[tex]Passed: 14[/tex]

So, the fraction of those that passed is:

[tex]Fraction = \frac{Passed}{n}[/tex]

[tex]Fraction = \frac{14}{20}[/tex]

Reduce fraction

[tex]Fraction = \frac{7}{10}[/tex]

Answer:

it's called the horizontal line

Step-by-step explanation:

also known as y=0

Answer:

12

Step-by-step explanation:

Answer:

3.5

Step-by-step explanation:

It's technically 3.69 but 3.5 is the closest answer you can get

Answer:

Part A

The volume of the triangular prism is 500 cm³

Part B

The total surface area of the prism is approximately 441.42 cm²

Step-by-step explanation:

The given details are;

The dimensions of the side length of the cube, s = 10 cm

The shape the cube was cut across to obtain = A prism

Part A

Whereby the prism obtained is a triangular prism, we have;

The cube can be cut in half to form a triangular prism

The volume of each triangular prism obtained = (1/2) × The volume of the cube

∴ The volume of the triangular prism = (1/2) × (10 cm)³ = 500 cm³

Part B

The height of the prism, h = 10 cm × sin(45°) = 5·√2 cm = (1/2) × The base width of the prism

The triangular cross sectional area of the prism, A₁ = 5·√2 × 5·√2 = 50

The square cross sectional area, A₂  = 10 × 10 = 100

The cross sectional area of the base, A₃ = 10·√2 × 10 = 100·√2

The total surface area of the prism, A = 2·A₁ + 2·A₂ + A₃

∴ A = 2×50 + 2×100 + 100·√2 = 300 + 100·√2 ≈ 441.42

The total surface area of the prism, A ≈ 441.42 cm²

Answer: The answer is the diameter.

Step-by-step explanation:

Answer:

diameter

Step-by-step explanation:

the diameter goes straight through the center of the circle

Answer:

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

Step-by-step explanation:

[tex]tan^ 2 \theta - ( \sqrt 3 + \frac{1}{\sqrt3}}) tan \theta + 1 = 0\\\\tan \theta - ( \sqrt 3 + \frac{1}{\sqrt3}}) +\frac{1}{ tan \theta } = 0\\\\[/tex]          [tex][ \ divide \ by \ tan \theta \ on \ both \ sides \ ][/tex]

[tex]tan\theta + \frac{1}{ tan \theta }- ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\\frac{tan^2 \theta + 1}{ tan \theta } - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\\frac{sec ^2 \theta}{ \frac{sin \theta }{cos \theta}} - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0[/tex]              [tex][ \tan ^ 2\theta + 1 = sec ^2 \theta \ , \ tan \theta = \frac{sin \theta }{cos \theta } \ ][/tex]

[tex]\frac{sec^2 \theta }{sin \theta \times sec \theta } - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\[/tex]                  [tex][\ \frac{sin \theta }{cos \theta } = sin \theta \times sec \theta \ ][/tex]

[tex]\frac{sec \theta }{sin \theta } - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\[/tex]

[tex]sec \theta \ cosec \theta - ( \sqrt 3 + \frac{1}{\sqrt3}}) = 0\\\\[/tex]              [tex][ \ \frac{1}{sin \theta } = cosec \theta \ , \ \frac{ sec \theta }{sin \theta } = sec \theta cosec \theta \ ][/tex]

[tex]sec \theta \ cosec \theta - \sqrt 3 - \frac{1}{\sqrt3}} = 0\\\\\frac{\sqrt 3\ sec \theta \ cosec \theta - 3 - 1}{\sqrt3} = 0\\\\\sqrt 3 sec \theta cosec \theta - 4 = 0\\\\[/tex]              

[tex]\sqrt3 \frac{1}{cos \theta } \frac{1}{sin \theta } - 4 = 0\\\\\frac{\sqrt3 - 4sin \theta cos \theta} { sin \theta cos \theta } = 0[/tex]                      

[tex]\sqrt 3 - 2sin 2\theta = 0[/tex]                                  [tex][ \ sin 2 \theta = 2 sin \theta cos \theta \ ][/tex]

[tex]2sin 2 \theta = \sqrt3\\\\sin 2 \theta = \frac{\sqrt3 }{2} \\\\2 \theta = sin^{-1} (\frac{\sqrt3}{2})\\\\2 \theta = 60^{ \circ} = \frac{ \pi}{3}\\\\\theta = \frac{\pi} {6}[/tex]

Answer:

3  7/15

Step-by-step explanation:

1 4/5 + 2 2/3 - 16/15

9/5 + 8/3 - 16/15

27/15 + 40/15 - 16/15

52/15 = 3  7/15

Answer:

3 2/5

Step-by-step explanation:

STEP 1: Convert all fractions into improper fractions

9/5 + 8/3 - 16/15

Step 2: pass denominators

(9x3 + 8x5 - 16) / 15

=51/15

Step it into proper mixed number

3 2/5

Answer:

A pair of linear functions

Step-by-step explanation:

Given

See comment for options

Required

Which has real numbers as its range and domain

The answer to this is, the linear functions

Take for instance

[tex]f(x) = mx + b[/tex]

The above function is real for all values of x

The inverse is calculated as thus:

[tex]f(x) = mx + b[/tex]

Replace f(x) with y

[tex]y = mx + b[/tex]

Swap y and x

[tex]x = my + b[/tex]

Make y the subject

[tex]my = x - b[/tex]

Divide by m

[tex]y = \frac{x - b}{m}[/tex]

Replace y with the inverse function

[tex]f^{-1}(x) = \frac{x - b}{m}[/tex]

The above function (i.e. the inverse function) is real for all values of x

This is not true for other options

Answer:

He made the following mistake, he assumed that polynomial [tex](x^{2}+1) = (x^{2}-1)[/tex], having for granted that [tex]x^{2}+1[/tex] has two real roots, instead of two complex roots.

Step-by-step explanation:

He made the following mistake, he assumed that polynomial [tex](x^{2}+1) = (x^{2}-1)[/tex], having for granted that [tex]x^{2}+1[/tex] has two real roots, instead of two complex roots. The true factorized form of the fourth grade polynomial is:

[tex]r(x) = (x^{2}+1)\cdot (x^{2}-9)[/tex]

[tex]r(x) = (x- i)\cdot (x+i)\cdot (x+3)\cdot (x-3)[/tex]

Answer:

B) 376 units squared

Step-by-step explanation:

Imma put a picture that has a fully explanation. Hope this helps, good luck! ;)

Btw saw your other one got deleted •_• *sigh*

Answer:gotta go sorry

Step-by-step explanation:

Answer:

13[tex]\sqrt{22}[/tex] / 44

Step-by-step explanation:

cos A = 9/13

here adjacent = 9 and hypotenuse = 13  opposite = ?

using pythagoras theorem to find opposite

a^2 + b^2 = c^2

9^2 + b^2 = 13^2

81 + b^2 = 169

b^2 = 169 - 81

b^2 = 88

b = [tex]\sqrt{88}[/tex]

b =  [tex]2\sqrt{22}[/tex]

therefore opposite =  [tex]2\sqrt{22}[/tex]

cosec A = hypotenuse/opposite

= 13/[tex]2\sqrt{22}[/tex]

rationalizing the denominator

=13/ [tex]2\sqrt{22}[/tex]  *  [tex]2\sqrt{22}[/tex] /  [tex]2\sqrt{22}[/tex]

=13 *[tex]2\sqrt{22}[/tex] /( [tex]2\sqrt{22}[/tex] )^2

=26 [tex]\sqrt{22}[/tex] / 4*22

=26 [tex]\sqrt{22}[/tex] / 88

=13[tex]\sqrt{22}[/tex] / 44

Answer:

[tex]cosec A = \frac{13}{\sqrt{88}}[/tex]

   OR

[tex]cosec A = \frac{13 \sqrt{22}}{44}[/tex]

Step-by-step explanation:

Formulas used:

[tex]cos^2 A = 1 - sin^2 A\\\\cosec A = \frac{1}{sin A}[/tex]

Given :

  [tex]cos A = \frac{9}{13}[/tex]

Find cosec A

   [tex]sin ^2 A = 1 - cos^2 A[/tex]

              [tex]= 1 - (\frac{9}{13})^2\\\\= 1 - \frac{81}{169}\\\\=\frac{169 - 81}{169}\\\\=\frac{88}{169}[/tex]

 [tex]sin A = \sqrt{\frac{88}{169}} = \frac{\sqrt{88}}{13}[/tex]

Therefore,

         [tex]cosec A = \frac{1}{sin A} = \frac{1}{ \frac{\sqrt{88}}{13}} = \frac{13}{ \sqrt{88}}[/tex]

OR In a simplified form :

[tex]cosec A = \frac{13}{\sqrt{88}} \times \frac{\sqrt{88}}{\sqrt{88}} = \frac {13 \times \sqrt{4 \times 22}}{88} = \frac{13 \times 2 \sqrt{22}} {88} = \frac{13 \sqrt{22}}{44}[/tex]        

Answer:

I think you mean significant figures

Step-by-step explanation:

Answer:

Range: 0.07

Median: 0.145

Mode: none

Step-by-step explanation:

Range: largest #- smallest #

Median: middle number (when numbers are in order least to greatest) or when two are middle number you add them, and divide by two

Mode: most occurring number(s) (if there aren't any, no mode)—but if there are multiples such as 2 zeros and 2 ones then 0 and 1 would be the mode