Determine the period.
N
10

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

Simplify,

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

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

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer:

No, they cannot. Some polynomials are different than others and can be factored in different ways. Some can't even be factored at all.

Perfect Square Trinomials have a pattern with factoring and can be written as one binomial squared.

Difference of Squares has a pattern of (x - y)(x + y).

Difference/Sum of Cubes have formulas for their patterns.

Prime Polynomials simply cannot be factored.

Step-by-step explanation:

Brainliest, please! (Almost an Ace)

Answer:

AB

Step-by-step explanation:

The segments that are parallel need to be in the same direction ( up and down)

The segments that are parallel are FH, AB, GC

Answer:

AB

Step-by-step explanation:

since is a cube all of the angles are 90 degees and this only possibel whn the line That a vertical a parrelllt to each other

Answer:

-117 is irrational number

Answer:

Irrational

Step-by-step explanation:

Irrational number can't be written as a faction, -11pie can't be written as a fraction. Therefore it is a irrational number.

Answer:

33/14

Step-by-step explanation:

[tex] - 10 \frac{2}{7} + ( - 4 \frac{4}{11} )[/tex]

[tex] = - \frac{72}{7} \div - ( \frac{48}{11} )[/tex]

[tex] = \frac{72}{7} \times \frac{11}{48} [/tex]

[tex] = \frac{3}{7} \times \frac{11}{2} [/tex]

[tex] = \frac{33}{14} [/tex]

stan dreamcatcher

Hi there!

[tex]\large\boxed{(-3, 8.25)}[/tex]

Solve for the vertex by completing the square:

y = -0.25x² - 1.5x + 6

Factor out a negative:

y = -(0.25x² + 1.5x) + 6

Remember that a square binomial is:

a² + 2ab + b²

We know that:

a² = 0.25, so a = 0.5

1.5 = 2ab, so:

1.5 = 2(0.5)b

b = 1.5

b² = 2.25

Thus:

y = -(0.25x² + 1.5x + 2.25) + 6 - (-2.25)

Simplify:

y = -(0.5x + 1.5)² + 8.25

Find the vertex by factoring out 0.5:

y = -0.5(x + 3)² + 8.25

Thus, the vertex is:

(-3, 8.25)

Answer:

1/25

Step-by-step explanation:

f(x) = 5^x

Let x = -2

f(-2) = 5^-2

We know a^-b = 1/a^b

f(-2) = 1/5^2

      = 1/25

Answer:

[tex]\frac{1}{25} [/tex]

Step-by-step explanation:

[tex]f(x) = {5}^{x} \\ f( - 2) \\ f( - 2) = {5}^{ - 2} \\ = \frac{1}{ {5}^{2} } \\ = \frac{1}{25} [/tex]

Answer:

I found x for you, right there.

Answer:  x is 12 and y is 5

Step-by-step explanation:

Answer:

1/2

Step-by-step explanation:

The radius is 1/2 of the diameter

a radius is 1/2 the diameter

Answer:

{1, 0, 16}

Step-by-step explanation:

given..

j(x) = x^2-2x+1

put all given values of domain (1,0 and 5 ) in the equation..

the values you get are range of the function

Answer: 1

Step-by-step explanation

0(2)-2(0)+1=0+0+1= 1

1(1)-(2)(1)+1=1-2+1= 1

(5)(2)-(2)(5)+1=10-10+1= 1

Step-by-step explanationIf you are reading this say

thank u