Do the activity please it's trigonometry ​

ANS=40

hope this help you

bye have a great day :)

Answer:

17%

Step-by-step explanation:

Again, as the amount of years increase, the population of bees gets multiplied by 0.83. We can rewrite this to 83%, and then again rewrite this to 100%-17%. We can see now that the population of bees decreases by 17% each year.

Answer:

0.1009

Step-by-step explanation:

To convert percentage into decimal, you need to divide the percentage by 100

10.09/100

= 0.1009

Answer:

13

Step-by-step explanation:

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: 14x^5 + 17x^4 - 87x^3 + 60x^2 + 15x - 18

Step-by-step explanation: You would need to simplify the expression by using distributive property.

Answer:

9 years older

Step-by-step explanation:

The ratio of their ages is 8 : 5 = 8x : 5x ( x is a multiplier )

In 30 years their ages will be 8x + 30 and 5x + 30 and the ratio 6 : 5 , so

[tex]\frac{8x+30}{5x+30}[/tex] = [tex]\frac{6}{5}[/tex] ( cross- multiply )

5(8x + 30) = 6(5x + 30) ← distribute parenthesis on both sides

40x + 150 = 30x + 180 ( subtract 30x from both sides )

10x + 150 = 180 ( subtract 150 from both sides )

10x = 30 ( divide both sides by 10 )

x = 3

Then

Michell is 8x = 8 × 3 = 24 years old

Connor is 5x = 5 × 3 = 15 years old

Mitchell is 24 - 15 = 9 years older than Connor

Answer:

sqrt(3) /3

Step-by-step explanation:

1 / sqrt(3)

Multiply the top and bottom by sqrt(3)

1/ sqrt(3) * sqrt(3)/ sqrt(3)

sqrt(3) /  sqrt(3)*sqrt(3)

sqrt(3) /3

Answer:

[tex] = { \sf{ \frac{1}{ \sqrt{3} } }} \\ \\ { \sf{ = \frac{1}{ \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} } }} \\ \\ = { \sf{ \frac{ \sqrt{3} }{ {( \sqrt{3}) }^{2} } = \frac{ \sqrt{3} }{3} }} [/tex]

Answer:

228573

Step-by-step explanation:

a = 219 (first term)

an = 2018 (last term)

Sn->Sum of n terms

Sn=n/2(a + an)         [Where n is no. of terms] -> eq 1

To find number of terms,

an = a + (n-1)d     [d->Common Difference] -> eq 2

d= 226-219 = 7

=> d=7

Substituting in eq 2,

2018 = 219 + (n-1)(7)

1799 = (n-1)(7)

1799 = 7n-7

1799 = 7(n-1)

1799/7 = n-1

257 = n-1

n=258

Substituting values in eq 1,

Sn = 258/2(219+2018)

    = 129(2237)

    = 228573

Answer:

Ekta and Preyal

Step-by-step explanation:

[tex]\\ \rm\Rrightarrow -30=5(x+1)[/tex]

[tex]\\ \rm\Rrightarrow -30=5x+5[/tex]

[tex]\\ \rm\Rrightarrow 5x=-30-5[/tex]

[tex]\\ \rm\Rrightarrow 5x=-35[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{-35}{-5}[/tex]

[tex]\\ \rm\Rrightarrow x=7[/tex]

Answer:

x = -7

Step-by-step explanation:

-30 = 5 (x -1 )

5 ( x + 1 ) =-30

5 (x + 1 ) = - 30

     5            5

x + 1 = -6

x + 1 -1 = -6 -1

x = - 7

Answer:

abc is a triangle so ,

a is ( 9,6 )

b is ( 9,3 )

and c is ( 3,3 )

Answer:

11

Step-by-step explanation:

6+(2x+28)=x+23=-11 -> Answer can't be negative -> 11

Answer:

divide, 2x+9

Step-by-step explanation:

got it right

Answer:

Step-by-step explanation:

Answer:

your answer is 30

Step-by-step explanation:

I hope this help

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

[tex]\\ \sf\longmapsto m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]\\ \sf\longmapsto m=\dfrac{1-3}{-4-3}[/tex]

[tex]\\ \sf\longmapsto m=\dfrac{-2}{-7}[/tex]

[tex]\\ \sf\longmapsto m=\dfrac{2}{7}[/tex]

Points are (-7,6),(11,-4)

[tex]\boxed{\sf slope(m)=\dfrac{y_2-y_1}{x_2-x_1}}[/tex]

[tex]\\ \sf\longmapsto m=\dfrac{-4-6}{11+7}[/tex]

[tex]\\ \sf\longmapsto m=\dfrac{-10}{18}[/tex]

[tex]\\ \sf\longmapsto m=-\dfrac{5}{9}[/tex]

Answer:

Step-by-step explanation:

Slope = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

9) Mark any two point on the line

(x₁ , y₁) = (3 , 3)   ;   (x₂, y₂) = (-4 ,1)

[tex]Slope =\frac{1-3}{-4-3}\\\\=\frac{-2}{-7}\\\\=\frac{2}{7}[/tex]

10) (x₁ , y₁) = ( -7 , 6)   ;   (x₂, y₂) = (11 ,-4)

[tex]Slope =\frac{-4-6}{11-[-7]}\\\\ =\frac{-4-6}{11+7}\\\\=\frac{-10}{18}\\\\=\frac{-5}{9}[/tex]

Answer:

-1

Step-by-step explanation:

I know that i^4 = 1

i^10 = i^4 * i^4 * i^2

     = 1 * 1 * i^2

We know that i^2 = -1

    =1 *1 *-1

    = -1

The type of budget that would likely work best for Brody is biweeky budget.

Budget is an economic term that refers to the planning and advance formulation of expenses and income. The budget is a tool to organize expenses depending on the amount of money available.

The type of budget that would be best for Brody is a biweekly budget because he receives his payment every fifteen days (twice a month). So, he  can schedule his expenses each time he receives his payment, in this way he  does not spend all his money before he receives the next payment.

Additionally, weekly, monthly, and dairy are not correct options because they do not fit the time periods in which Brody receives payment for his services.

Learn more in: https://brainly.com/question/141889

Note:

This question is incomplete because options are missing, here are the options.

Daily budget

Biweekly budget

Monthly budget

Weekly budget

Step-by-step explanation:

Given that,

We have to find the value of m∠E.

Here, two sides are equal, thus it is an isosceles triangle. As the two sides are equal, so their angles must be equal. So, ∠E and ∠D will be equal. Let us assume the measures of both ∠E and ∠D as x.

→ Sum of all the interior angles of ∆ = 180°

→ ∠E + ∠D + ∠F = 180°

→ 116° + x + x = 180°

→ 2x = 180° – 116°

→ 2x = 64°

→ x = 64° ÷ 2

→ x = 32°

Henceforth,

→ m∠E = x

→ m∠E = 32°

[tex] \\ [/tex]

~

Answer:

Step-by-step explanation:

Question:

Max needs to purchase a car and  withdraws $100 from his bank. In a few days he withdraws another $50 to make same repairs. In total what is the change in his bank balance from theese two costs?

Solution:

(-100) + (-50) =

-150

Answered by G a u t h m a t h

Answer:

günah(3x) + 2

Step-by-step explanation:

Gösterilen sinüzoidal fonksiyonun denklemini yazınız? A) y = cos x + 2 B) y = cos(3x) + 2 C) y = günah x + 2 D) y =

Answer:

y = sin(3x) + 2

Answer:

4+2x = 12

Step-by-step explanation:

sum means add an is means equal

4+2x = 12

Step-by-step explanation:

the sum of 4 and twice a number is 12:

Answer:4 in each bad 2 left over

Step-by-step explanation:

Answer:

4 in each bag and 2 left over

Step-by-step explanation:

divide 14 by 3

3 goes into 14, 4 times

14 - 12 = 2

4 in each bag and then 2 left over

3x+x^2=3x+x^2

3x-3x=x^2-x^2

which means x=0

Step-by-step explanation:

A natural number is a positive whole number.

A whole number is a positive number with no fractions or decimals.

A interger is a whole number negative or positive.

A rational number is a number that terminates or continue with repeating digits.

A irrational number is a number that doesn't terminate or continue with repeating digits.

1. Rational Number

2. Natural,Whole,Interger,Rational

3. Whole,Rational,Interger

4. Rational

5.Irrational

6.Rational

7.Natural,Whole,Interger,Rational

8.Interger,Rational

9.Irrational

Answer:

first odd integer=x

second odd integer=x+2

third odd integer=x+4

x+x+2+x+4=2097

x+x+x+2+4=2097

3x+6=2097

3x=2097-6

3x=2091

3x/3=2091/3

x=697

therefore, x=697

x+2=697+2=699

x+4=697+4=701

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]