[100 points]
Using mathematically precise language, explain in detail how you would multiply the complex number z1=r1 (cos theta1+isin theta 1) with the complex number z2=r2 (cos theta 2+isin theta 2)

Answer:

$272

Step-by-step explanation:

9x7x12.5=787.5 120x7=840  38.95x7= 272.65 and she said to round to the nearest dollar so $272 dollars

Step-by-step explanation:

It's not 3 triangles and 1 square because you need four triangles to make the siding and the square wont go anywhere because of how it's going to look.

It's not 3 triangles and 1 rectangle because still you need four to make the sides and a rectangle is too long.

4 triangles and 1 rectangle is still not correct because the rectangle on the bottom of the net is going to mess it up fir the whole shape.

So in conclusion, 4 triangles and 1 squares is the correct answer.

Answer:

[tex]x {}^{2} - 36 = 0 \\ \\ x {}^{2} = 36 \\ \\ x = \sqrt{36} \\ \\ x = + 36 \: \: or \: \: - 36[/tex]

hope helpful -,-

Hey there!

3x = 27
DIVIDE 3 to BOTH SIDES

3x/3 = 27/3

CANCEL: 3/3 because it give you 1

KEEP: 27/3 because it help solve for the x-value

NEW EQUATION: x = 27/3

SIMPLIFY IT!

x = 9

Good luck on assignment & enjoy your day!

~Amphitrite1040:)

{1, 3, 5, 7, 9, 11, ... , 47}

To Write The Following in Set-builder notation

[tex] \text{A} = [/tex] {x : x ∈ N;x is an odd number, x ≤ 47}

Answer:

if not 24.5, it should be 31 yds

Step-by-step explanation:

ok so, I don't know how you want me to solve this but, if you're asking the distance diagonally, then it will be 24.5 yds. If you want the distance in straight lines, it will be 31 yds

lol there wasn't much information, sorry

Consider the numbers as, if we assume our first number to be x and second number being y, then from the given information we will be having the following equations :

[tex]{:\implies \quad \begin{cases}\sf 3x+3y=-12\\ \\ \sf 5x+2y=4\end{cases}}[/tex]

Now, rewrite the second equation as ;

[tex]{:\implies \quad \sf 3(x+y)=-12}[/tex]

Divide both sides by 3 ;

[tex]{:\implies \quad \sf x+y=-4\quad ---(i)}[/tex]

Now, consider the second equation of the above cases

[tex]{:\implies \quad \sf 5x+2y=4\quad ---(ii)}[/tex]

Now, multiply (i) by 2 on both sides :

[tex]{:\implies \quad \sf 2x+2y=-8\quad ---(iii)}[/tex]

Now, subtracting (ii) from (iii) will give us :

[tex]{:\implies \quad \sf 2x+2y-(5x+2y)=-8-4}[/tex]

[tex]{:\implies \quad \sf 2x+2y-5x-2y=-12}[/tex]

[tex]{:\implies \quad \sf -3x=-12}[/tex]

[tex]{:\implies \quad \boxed{\bf{x=4}}}[/tex]

Now, putting x = 4 in (i), will give y + 4 = -4, then solving for y will yield [tex]{\boxed{\bf{y=-8}}}[/tex]

Hence, we can conclude that :

[tex]{\quad \qquad \longrightarrow \begin{cases}\bf x=4\\ \\ \bf y=-8\end{cases}}[/tex]

Shaded area = area of square - area of a circle.

[tex] \\ \\ [/tex]

For this we have to find area of both square and circle which is visible in the picture.

[tex] \\ [/tex]

Part one :

In this part we'll find area of circle.

Given :

To find :

Solution:

We know :

[tex] \boxed{ \rm Area \: of \: circle =\pi {r}^{2} }[/tex]

Steps :

[tex] \dashrightarrow\sf Area \: of \: circle =\pi {r}^{2}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: circle = \dfrac{22}{7} \times {9}^{2} [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: circle = \dfrac{22}{7} \times 9 \times 9[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: circle = \dfrac{22}{7} \times 81[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: circle = \dfrac{22\times 81}{7}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: circle = \dfrac{1782}{7}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\bf Area \: of \: circle =254.57 \: ft {}^{2} \\ [/tex]

[tex] \\ [/tex]

Part 2

As it's visible diameter of circle = side of square.

So we have to find diameter of circle.

We know :-

[tex] \boxed{ \rm Diameter \: of \: circle =2 \: radius}[/tex]

[tex] \\ [/tex]

Steps:-

[tex] \dashrightarrow\sf Diameter \: of \: circle =2 \: radius \\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Diameter \: of \: circle =2 \times 9\\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\bf Diameter \: of \: circle =18 \: ft[/tex]

We have :

To find:

Solution:

We know :

[tex] \boxed{ \rm Area \: of \: square= {side}^{2} }[/tex]

Steps :

[tex] \dashrightarrow\sf Area \: of \: square= {side}^{2}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: square= {18}^{2}[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\sf Area \: of \: square= {18} \times 18[/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\bf Area \: of \: square= 324 \: {ft}^{2} [/tex]

Part three:

Remember the first line of answer ? :)

So let's insert here:-

[tex] \boxed{ \text{Shaded area = area of square - area of a circle}}[/tex]

Steps :-

[tex] \dashrightarrow\textsf{Shaded area = area of square - area of a circle} \\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\textsf{Shaded area = 324 - 254.57} \\ [/tex]

[tex] \\ [/tex]

[tex] \dashrightarrow\textbf{Shaded area = 69.43 }\bf {ft}^{2} \\ [/tex]

Answer:

b is equal to 14 or higher

Step-by-step explanation:

Answer:

convert the mixed fractions into decimals. do this by converting them into improper fractions and dividing the numerator by the denominator

Answer:

∠ 1 = 155° , ∠ 2 = 25°

Step-by-step explanation:

the 2 angles are adjacent angles and sum to 180° , that is

5x + x - 6 = 180

6x - 6 = 180 ( add 6 to both sides )

6x = 186 ( divide both sides by 6 )

x = 31

then

∠ 1 = 5x = 5(31) = 155°

∠ 2 = x - 6 = 31 - 6 = 25°

Answer:

-940.5

Step-by-step explanation:

Answer:

-940.5

Step-by-step explanation:

use pemdas

1.parenthesis

2.exponets

3.multiply

4.divide

5.add

6.subtract

Answer:

x = 2

Step-by-step explanation:

24 - 7x = 10

Rearrange terms,

-7x + 24 = 10

Subtract 24 from both sides,

-7x + 24 - 24 = 10 - 24

-7x = 10 - 24

-7x = -14

Divide by sides by -7,

[tex]\frac{-7x}{-7}=\frac{-14}{-7}[/tex]

[tex]x=\frac{-14}{-7}[/tex]

x = 2

Answer: This is wrong. It takes 4 men to dig a hole and you cannot dig half a hole.

Step-by-step explanation:

Answer:

u have to be kidding me 2 Men can not dig a hole it takes like 4 or 5 men

Answer:

34.5 cm²

Step-by-step explanation:

The area of the shaded region is the area of the rectangle minus the area of the right triangle.

The area of the rectangle is simply the length 7 cm multiplied by the width 6 cm.

To find the area of the right triangle, you must first find the base and height of the triangle.

The two lengths can be found this way:

7 cm - 2 cm = 5 cm

6 cm - 3 cm = 3 cm

area of shaded region = LW - bh/2

area of shaded region = 7 cm × 6 cm - 5 cm × 3 cm / 2

area of shaded region = 42 cm² - 7.5 cm²

area of shaded region = 34.5 cm²

Answer:

263.86cm²

Step-by-step explanation:

75/4 = 18.75

to find width = 18.75×1.5=28.13

to find length = 18.75×0.5=9.38

28.13×9.38=263.86cm²

180° hope it's correct and really helps....!!!!

Answer:

64

Step-by-step explanation:

y = x + 4

y - x = 4

4^3 = 64

the will he worth less because 80% a year is a 10% so 2600-500=2100 so 2100÷10=210

Answer:

m∠F = 100

Step-by-step explanation:

Sum of interior angles in a hexagon is 720.

( x - 60 ) + ( x - 40 ) + 130 + 120 + 110 + ( x - 20 ) = 720

x + x + x - 60 - 40 - 20 + 130 + 120 + 110 = 720

3x - 120 + 120 + 130 + 110 = 720

3x + 130 + 110 = 720

3x + 240 = 720

3x = 720 - 240

3x = 480

x = 480 / 3

x = 120

m∠F = x - 20

Substitute the value of x,

m∠F = 120 - 20

m∠F = 100

Answer:

415ml

Step-by-step explanation:

there is 1000 ml in 1litre so the question says half which means 500 ml so if he drank 85ml then

500ml-85ml= 415 ml

The total number of gallons of diesel Truck B used to travel 132 miles is 11.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

A dairy farmer uses two trucks to deliver milk. The two trucks use different kinds of fuel. Truck A uses gasoline and Truck B uses diesel.

From the given table we can see that

The total number of gallons of diesel Truck B used to travel 12 miles = 1

Number of gallons of diesel Truck B used to travel 1 mile = 1 /12

The total number of gallons of diesel Truck B used to travel 132 miles

= 1 /12 x 132

= 132 / 12

= 11

Thus, The total number of gallons of diesel Truck B used to travel 132 miles is 11.

Learn more about the unitary method;

https://brainly.com/question/23423168

#SPJ1

Answer:

  278°

Step-by-step explanation:

You want the measure of arc RSQ given inscribed angle RSQ has a measure of 41°.

An inscribed angle has half the measure of the arc it subtends. This means short arc RQ will have double the measure of angle RSQ:

  arc RQ = 2 × 41° = 82°

The measure of long arc RSQ will complete the 360° circle:

  arc RSQ = 360° -arc RQ

  arc RSQ = 360° -82°

  arc RSQ = 278°

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Answer:

50

Step-by-step explanation:

It is on edge

Answer:

it is counting 48 than going down

Step-by-step explanation:1,48

[tex]\displaystyle \int (\sin 2x + \cos 2x)~ dx\\\\=\displaystyle \int \sin 2x ~dx + \displaystyle \int \cos 2x ~ dx\\\\=-\dfrac{\cos 2x}2 + \dfrac{\sin 2x}2 +C\\\\=\dfrac{\sin 2x - \cos 2x}2+C[/tex]

Answer:

10 men can complete work l in 120 doy . In how many days does 30 me can complete it.

-

Answer: [tex]y=Ce^(^3^t^{^9}^)[/tex]

Step-by-step explanation:

Beginning with the first differential equation:

[tex]\frac{dy}{dt} =27t^8y[/tex]

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

[tex]\frac{1}{y} \frac{dy}{dt} =27t^8[/tex]

Multiply both sides by 'dt' to get:

[tex]\frac{1}{y}dy =27t^8dt[/tex]

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

[tex]\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt[/tex]

[tex]ln(y)=27(\frac{1}{9} t^9)+C[/tex]

[tex]ln(y)=3t^9+C[/tex]

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

[tex]e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)[/tex]

[tex]y=e^(^3^t^{^9} ^+^C^)[/tex]

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

[tex]y=e^(^3^t^{^9}^)e^C[/tex]

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

[tex]y=Ce^(^3^t^{^9}^)[/tex]

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

[tex]\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))[/tex]

[tex]\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)[/tex]

[tex]\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8[/tex]

Now check if the derivative equals the right side of the original differential equation:

[tex](Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)[/tex]

[tex]Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)[/tex]

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.