Solve 4 sinx + 9 cosx=0 for 0° ​

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:

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:

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

a) 26 x (-48) + (-48) x (-36) = ( –1248) + ( + 1728) = – 1248+ 1728 = 480

b) 625 x (-35) + (-625) x 65 = ( –21875) + ( –40625) = – 21875 –40625 = –62500

I hope I helped you^_^

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

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

-- (central angles)

So, 35 + 65 = 100 degrees

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]

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!

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:

7t + 6 + 9v

Step-by-step explanation:

7t + 6 + 3v + 6v (since 3v and 6v are like terms you will add them both.)

7t + 6 + 9v

Hope this helps, thank you :) !!

Answer:

7t+6+9v

Step-by-step explanation:

7t+6+3v+6v

7t has no opponent it is =7t

6 is on it own =6

3v+6v=9v,reason is 3v has an opponent which is 6v so addition of 3v and 6v is =9v

so ur ans. is =7t+6+9v

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:

(-3)

Step-by-step explanation:

follow me if you want

Answer:

hope it help you

Step-by-step explanation:

mark me brailiest answer

Answer:

Hello,

[tex]\begin{bmatrix}\dfrac{-1}{5} &0\\\dfrac{-1}{10}&\dfrac{1}{4}\end{bmatrix}[/tex]

Step-by-step explanation:

See jointed file

The inverse of the matrix A is

[tex]A = \left[\begin{array}{cc}-5&-2\\0&4\end{array}\right][/tex]

A matrix is a set of numbers arranges in rows and columns such that it form a rectangular array.

We have,

[tex]A = \left[\begin{array}{cc}-5&0\\-2&4\end{array}\right][/tex]

The matrix A is a square matrix and its determinant is not zero.

So the inverse exist.

The inverse of matrix A.

We will change the rows into columns and columns into rows.

So,

[tex]A = \left[\begin{array}{cc}-5&-2\\0&4\end{array}\right][/tex]

Thus,

The inverse of the matrix A is given above.

Learn more about matrix here:

https://brainly.com/question/28180105

#SPJ2

Answer:

47.746, answer choice C

Step-by-step explanation:

47.746

Answer: g(x) = (1/2)3^-x  reflection over y axis yields (-x,y)

Answer:

divide, 2x+9

Step-by-step explanation:

got it right

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

9514 1404 393

Answer:

  equals can be substituted anytime anywhere

Step-by-step explanation:

cos(60°) = 1/2, so wherever one appears, the other can be substituted. This is allowed by the substitution property of equality.

__

If you don't substitute at some point, you find the answer to be ...

  x = 10/cos(60°)

Most of us are interested in a numerical value for x, so we prefer that cos(60°) be replaced by a numerical value.

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:

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]

Answer:

x = 100°

Step-by-step explanation:

p and q are parallel lines. Construct line 'l' parallel to p and q

a = 30°   {p// l  when parallel lines are intersected by transversal alternate interior angles are congruent}

c +110 = 180  {linear pair}

c = 180  - 110

c = 70°

b= c   {q // l  when parallel lines are intersected by transversal alternate interior angles are congruent}

b = 70°

x = a + b

x = 30° + 70°

x = 100°

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:

216

Step-by-step explanation:

Volume of a rectangular prism = area of base * depth

Area of base: 72

Depth: 3

Volume = 72 * 3 = 216