Julia earned $2,000 during the month of September.
Use the paycheck statement to find the missing percent rates of the taxes Julia payed.

Take RHS

[tex]\\ \rm\Rrightarrow \dfrac{tanA+tanB}{tanA-tanB}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{\dfrac{sinA}{cosA}+\dfrac{sinB}{cosB}}{\dfrac{sinA}{cosA}-\dfrac{sinB}{cosB}}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{\dfrac{sinAcosB+cosAsinB}{cosAcosB}}{\dfrac{sinAcosB-cosAsinB}{cosAcosB}}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sinAcosB+cosAsinB}{sinAcosB-cosAsinB}[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{sin(A+B)}{sin(A-B)}[/tex]

Some important identities

[tex]\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ $\: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1$\end{minipage}}[/tex]

Answer:

[tex]\boxed{\begin{minipage}{6 cm}\underline{Trigonometric Identities}\\\\$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B\\\\ \dfrac{\sin A}{\cos A}=\tan A$\\\end{minipage}}[/tex]

[tex]\dfrac{\sin (a+b)}{\sin (a-b)} & = \dfrac{\sin(a) \cos (b) + \cos (a) \sin (b)}{\sin(a) \cos (b) - \cos (a) \sin (b)}[/tex]

[tex]\dfrac{\sin (a+b)}{\sin (a-b)} & = \dfrac{\sin(a) \cos (b) + \cos (a) \sin (b)}{\sin(a) \cos (b) - \cos (a) \sin (b)} \times \dfrac{\dfrac{1}{\cos (a) \cos (b)}}{\dfrac{1}{\cos (a) \cos (b)}}[/tex]

[tex]\dfrac{\sin (a+b)}{\sin (a-b)} & = \dfrac{\dfrac{\sin(a) \cos (b)}{\cos (a) \cos (b)} + \dfrac{\cos (a) \sin (b)}{\cos (a) \cos (b)}}{\dfrac{\sin(a) \cos (b)}{\cos (a) \cos (b)} - \dfrac{\cos (a) \sin (b)}{\cos (a) \cos (b)}}[/tex]

[tex]\dfrac{\sin (a+b)}{\sin (a-b)} & = \dfrac{\dfrac{\sin(a)}{\cos (a)} + \dfrac{\sin (b)}{\cos (b)}}{\dfrac{\sin(a)}{\cos (a)} - \dfrac{\sin (b)}{\cos (b)}}[/tex]

[tex]\dfrac{\sin (a+b)}{\sin (a-b)}=\dfrac{\tan(a)+\tan(b)}{\tan(a)-\tan(b)}[/tex]

Answer:

a and d

or

c and b

Step-by-step explanation:

supplementary angles are angles that add up to 180 degrees

when looking at the rectangle you see two possible answers

Answer:

C = [tex]12\pi m[/tex]

A = [tex]36\pi m^2[/tex]

Step-by-step explanation:

I'm going to assume that the radius is 6m. The circumference of a circle is [tex]C = 2r\pi[/tex]. So we can plug in 6 for r and we get the circumference is [tex]12\pi m[/tex].

For area, the equation is [tex]A = \pi r^{2}[/tex]. So we can again plug in 6 for r and we get that the area is [tex]36\pi m^2[/tex].

If 6m is the diameter, then the circumference is [tex]6\pi m[/tex] and the area is [tex]9\pi m^2[/tex].

Answer:

Step-by-step explanation:

6 / 3√5

= 2/√5

= 2√5/5

Answer:

 [tex]\frac{2\sqrt{5} }{5}[/tex]

Step-by-step explanation:

First, you can take a factor of (3) out of the numerator and denominator. This simplifies to equal 1, so you are left with 3/√5.

6 / 3√5   ----->   (3/3)  x  2/1√5  --->  (1)  x  2/√5  --->  2/√5

Because radicals should not be kept in the denominator, multiply both the numerator and denominator by the radical of 5. This removes the radical from the denominator while altering both the numerator and denominator equally. Remember, the square root of a number is technically the number raised to the 1/2 power. The 1/2 power times the 1/2 power equals the power of one, essentially "eliminating" the radical.

2/√5  --->  2/√5  x  √5/√5  --->  2√5 / 5

As there is not way to simplify further, your final answer should be [tex]\frac{2\sqrt{5} }{5}[/tex].

Answer:

23

Step-by-step explanation:

23 ⇔ 23 R 0

184 divided by 8

=

23 with a remainder of 0

0

2

3

8

1

8

4

-

0

1

8

-

1

6

2

4

-

2

4

0

Answer: 2.99914cm Rounds up to 3

Step-by-step explanation:

Answer:

I dont think I can type the answer on keyboard

Answer:

1  5/8

Step-by-step explanation:

3/4 -  -7/8

3/4 + 7/8

6/8 + 7/8

13/8 = 1  5/8

Answer:

1 5/8 or 13/8

Step-by-step explanation:

Remove parentheses.

[tex]\frac{3}{4}+\frac{7}{8}[/tex]

Find the Least Common Denominator (LCD) of [tex]\frac{3}{4},\frac{7}{8}[/tex] . In other words, find the Least Common Multiple (LCM) of 4,8.

Method 1: By Listing Multiples

List the multiples of each number.

Multiples of 4 : 4, 8, ...

Multiples of 8 : 8, ...

Find the smallest number that is shared by all rows above. This is the LCM.

LCM = 8

Method 2: By Prime Factors

List the prime factors of each number

Prime Factors of 4 : 2, 2

Prime Factors of 8 : 2, 2, 2

Find the union of these primes.

2, 2, 2

Multiply these numbers: 2 * 2 * 2 = 8 ---> This is the LCM.

LCM = 8

---> LCD = 8

Make the denominators the same as the LCD.

[tex]\frac{3\times 2}{4\times 2}+\frac{7}{8}[/tex]

Simplify. Denominators are now the same.

[tex]\frac{6}{8}+\frac{7}{8}[/tex]

Join the denominators.

[tex]\frac{6+7}{8}[/tex]

Simplify.

[tex]\frac{13}{8}[/tex]

Convert to mixed fraction.

[tex]1\frac{5}{8}[/tex]

Answer:

D. Open the compass so that the distance from the two points of the compass is wider than half the length of [tex]\over{AB}[/tex].

Step-by-step explanation:

To construct a perpendicular for [tex]\over{AB}[/tex], we must first take a compass & take the distance of its arms wider than half the length of [tex]\over{AB}[/tex].

This is done in order to get two intersecting arcs in the top & bottom of [tex]\over{AB}[/tex] so that a perpendicular bisector can be drawn through it.

After two intersecting lines are drawn below & above [tex]\over{AB}[/tex], draw a line joining these 2 points through their points of intersection. The point where it intersects [tex]\over{AB}[/tex] is the middle-most point of [tex]\over{AB}[/tex] & now a perpendicular bisector of [tex]\over{AB}[/tex] is constructed.

[tex]\rule{150pt}{2pt}[/tex]

Answer:

D

Step-by-step explanation:

The trigonometric function gives the ratio of different sides of a right-angle triangle. The length of the side KL is 30.81 units.

The trigonometric function gives the ratio of different sides of a right-angle triangle.

[tex]\rm Sin \theta=\dfrac{Perpendicular}{Hypotenuse}\\\\\\Cos \theta=\dfrac{Base}{Hypotenuse}\\\\\\Tan \theta=\dfrac{Perpendicular}{Base}\\\\\\Cosec \theta=\dfrac{Hypotenuse}{Perpendicular}\\\\\\Sec \theta=\dfrac{Hypotenuse}{Base}\\\\\\Cot \theta=\dfrac{Base}{Perpendicular}\\\\\\[/tex]

where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.

The measure of ∠J can be written as,

Cos(∠J) = JK/JL

Cos(∠J) = 7/31.6

∠J = 77.2°

Similarly the length of side KL can be written as,

Tan(∠J) = KL/JK

Tan(77.2°) = KL/7

KL = 30.81 units

Hence, the length of the side KL is 30.81 units.

Learn more about Trigonometric functions:

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The property applied to the expression is (b) inverse property of multiplication

The expression is given as:

[tex]\frac{17}{3} * \frac{3}{17} = 1[/tex]

The numbers 17/3 and 3/17 are multiplicative inverses.

This means that the property applied to the expression is (b) inverse property of multiplication

Read more about expression properties at:

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Answer:

Inverse Multiplication

Step-by-step explanation:

If you want the work to be shown, look up, otherwise my answer was B on Edge 2022

Answer:

the cube root of 0.027 is 0.3

Answer:

last and third

Step-by-step explanation:

Answer:

a. 29.05 b. 29.05 c. 29.5

Answer:

9.435

Step-by-step explanation:

Answer:nine thousand six hundred and three

Step-by-step explanation:

Answer:

nine thousand six hundred and three

Step-by-step explanation:

Answer:

7. m^21n^27

8. a^21b^28

Step-by-step explanation:

7. since m^7 raised to the 3rd power = m^( 7 * 3 ) = m^21

then  n^9 raised to the 3rd power = n^( 9 * 3 ) = n^27

your answer is m^21n^27

8. a^3 raised to the 7th power = a( 3 * 7 ) = a^21

b^4 raised to the 7th power = b( 4 * 7 ) = b^28

your answer is a^21b^28

hope this helps:)

The simplified quotient of 300 and 3 is 100

The complete statement is given as:

What is the quotient of 300 and 3?

This is represented as:

Quotient = 300/3

Divide 300 by 3

Quotient = 100

Hence, the quotient of 300 and 3 is 100

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Answer:

(1,4) and (2,8)

Step-by-step explanation:

If you write these 2 ordered pairs as a fraction: 1/4 and 2/8 you can see that they are both equivalents.

Answer:

B, C and D

Step-by-step explanation:

Edge 2022/2023

Answer:

Step-by-step explanation:

Answer:

line

Step-by-step explanation:

quizlet

cliffnotes

The class that had better overall results on the exam is class A.

The class with greater variablity in results is class B.

The statistics of students in class A:

Maximum score = 100

Minimum score = 65

Median = 85

First quartile = 1/4(n + 1) = 1/4 x 12 = 3rd term = 70

Third quartile = 3/4 x (n + 1) = 3/4 x 12 = 9th term = 90

The statistics of students in class B:

Maximum score = 100

Minimum score = 55

Median = 75

First quartile = 70

Third quartile = 80

The range would be used to determine which class has greater variability.

Range = highest score  - lowest score

Class a = 100 - 65 = 35

Class B = 100 - 55 = 45

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Answer:

$10.64

Step-by-step explanation:

20% out of $53.18 is $10.636 rounded to $10.64

Answer:

i think it is D but I am not for shore

Answer:

she earns $1008 in commissions , hope this helps i did it

The image of triangle ABC is a A'B'C' after rotating about the origin with an angle of 180 degree.

It is defined as the change in coordinates and the shape of the geometrical body. It is also referred to as a two-dimensional transformation. In the geometric transformation, changes in the geometry can be possible by rotation, translation, reflection, and glide translation.

If we rotate the triangle ABC with the origin (0, 0) about 180 degree it will form an image of the triangle ABC which is A'B'C'

Here the coordinates for ABC are not given so we are assuming the triangle ABC has a image which is A'B'C' (refer attached picture)

Thus, the image of triangle ABC is a A'B'C' after rotating about the origin with an angle of 180 degree.

Learn more about the geometric transformation here:

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=========================================================

Explanation:

Let x = 0.370370370...

The "370" block repeats forever, due to the horizontal bar up top.

Multiply both sides by 1000 so that the decimal point moves three spots to the right.

x = 0.370370370...

1000x = 1000*0.370370370...

1000x = 370.370370370...

Then notice both x and 1000x involve the *exact* same decimal string of "370" repeating forever. This perfect match lets us cancel the decimal portions out when we subtract, and allows us to isolate x to get some fraction of integers.

1000x - x = 370.370370370... - 0.370370370...

999x = 370

x = 370/999

You can use your calculator to confirm that 370/999 = 0.370370370...

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In fact, you can use your calculator to determine the following:

A) 37/100 = 0.37

B) 370/999 = 0.370370370...

C) 37/99 = 0.373737...

D) 37/9 = 4.1111111....

This allows us to see fairly quickly that choice B is the final answer and that we can rule out the others.

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Side note: your calculator may round the last digit by mistake.