Mathematics College

Look at the figure:

An image of a right triangle is shown with an angle labeled x.

If tan x° = a divided by 4 and cos x° = 4 divided by b what is the value of sin x°?

sin x° = 4b
sin x° = b divided by a
sin x° = 4a
sin x° = a divided by b

Answers

Answer 1

By using trigonometric functions, the value of [tex]\sin \text{x}^\circ[/tex] is [tex]\frac{\text{a}}{\text{b}}[/tex].

What are trigonometric functions?

Trigonometric functions are also known as Circular Functions can be simply defined as the functions of an angle of a triangle. It means that the relationship between the angles and sides of a triangle are given by these trig functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant.

Given

[tex]\sin \text{x}^\circ= \ ?[/tex]

[tex]\tan \text{x}^\circ=\dfrac{\text{a}}{4}[/tex]

[tex]\cos \text{x}^\circ=\dfrac{4}{\text{b}}[/tex]

Formula to find [tex]\sin \text{x}^\circ[/tex]

[tex]\sin \text{x}^\circ=\dfrac{\text{Opposite}}{\text{Hypotenuse}}[/tex]

[tex]\rightarrow\sin \text{x}^\circ=\bold{\dfrac{a}{b}}[/tex]

Therefore, by using trigonometric functions, the value of [tex]\sin \text{x}^\circ[/tex] is [tex]\frac{\text{a}}{\text{b}}[/tex].

To know more about trigonometric functions, visit:

https://brainly.com/question/31540769

Look At The Figure:An Image Of A Right Triangle Is Shown With An Angle Labeled X.If Tan X = A Divided

Related Questions

Solve (9th grade math)

Answers

Answer:

1. Geometric Sequence Equation for Table A:

[tex]\boxed{\tt a_n = a_1 \times r^{(n-1)}}[/tex]

2. Arithmetic Sequence Equation for Table B:

[tex]\boxed{\tt a_n = a_1 + (n-1) \times d}[/tex]

3.

20th term for Geometric Sequence:
[tex]\boxed{\tt a_{20} = a_1 \times r^{19}}[/tex] 20th term for Geometric Sequence:
[tex]\boxed{\tt a_{20} = a_1 + 19\times d}[/tex]

Step-by-step explanation:

1. Geometric Sequence Equation for Table A:

The geometric sequence equation is given by the formula:

[tex]\tt \[a_n = a_1 \times r^{(n-1)}\][/tex]

where:

[tex]\tt \(a_n\)[/tex] represents the nth term of the sequence.[tex]\tt \(a_1\)[/tex] is the first term of the sequence.[tex]\tt \(r\)[/tex] is the common ratio.

For Table A, since we don't have the actual values, we can represent the equation as:

[tex]\boxed{\tt a_n = a_1 \times r^{(n-1)}}[/tex]

[tex]\hrulefill[/tex]

2. Arithmetic Sequence Equation for Table B:

The arithmetic sequence equation is given by the formula:

[tex]\tt a_n = a_1 + (n-1) \times d[/tex]

where:

[tex]\tt \(a_n\)[/tex] represents the nth term of the sequence.[tex]\tt \(a_1\)[/tex] is the first term of the sequence.[tex]\tt \(d\)[/tex] is the common difference.

For Table B, since we don't have the actual values, we can represent the equation as:

[tex]\boxed{\tt a_n = a_1 + (n-1) \times d}[/tex]

[tex]\hrulefill[/tex]

3. Finding Term 20 for both sequences:

In order to find the 20th term for both sequences, we need the actual values for [tex]\tt \(a_1\), \(r\)[/tex] and d.

in the case of Table A

[tex]\tt a_{20} = a_1 \times r^{(20-1)}[/tex]

[tex]\boxed{\tt a_{20} = a_1 \times r^{19}}[/tex]

in the case of Table B.

[tex]\tt a_{20} = a_1 + (20-1) \times d[/tex]

[tex]\boxed{\tt a_{20} = a_1 + 19\times d}[/tex]

By using this formula, we can easily fill up the box.

Find the original slope of (-6,-1) and (0,3)

Answers

The original slope would be -2/3

Answer:

slope (m) = 2/3

Step-by-step explanation:

slope = change in x /change in y

Also, slope is y2 - y1 / x2 -x1. That is what I apply for this activity, hence:

slope = 3 - (-1) / 0 - (-6)

         = 3 + 1 / 0 + 6

         = 4 / 6

         = 2/3

∴ slope(m) = 2/3

Differentiate the function. Then find an equation of the tangent line at the indicated point on the graph of the function.

Answers

Answer:

The derivative of the function:

[tex]f'(x)=-\dfrac{1}{2\sqrt{5-x} }[/tex]

The tangent line of the function at the given point:

[tex]y=-\dfrac{1}{6}x-\dfrac{25}{3}[/tex]

Step-by-step explanation:

Find the equation of the tangent line of the given function using the given point.

[tex]f(x)=6+\sqrt{5-x}; \ (x,y)=(-4,9)[/tex]

[tex]\hrulefill[/tex]

To find the equation of the tangent line to a function at a given point, follow these step-by-step instructions:

Step 1: Identify the point of tangency

Determine the x-coordinate of the point of tangency. Let's call it x₀.Find the corresponding y-coordinate of the point of tangency. Let's call it y₀.

Step 2: Find the derivative of the function

Calculate the derivative of the given function. Let's denote it as f'(x).

Step 3: Substitute the x-coordinate into the derivative

Replace the variable x in the derivative function f'(x) with the x-coordinate of the point of tangency (x₀).Evaluate the derivative at x₀ to find the slope of the tangent line. Let's denote it as m.

Step 4: Write the equation of the tangent line

Use the point-slope form of a line: y - y₀ = m(x - x₀).Substitute the values of m, x₀, and y₀ into the equation.Simplify and rearrange the equation to obtain the final form.

Step 5: Optional - Simplify the equation

If necessary, simplify the equation by performing any algebraic manipulations.

Step 6: Optional - Verify the equation

Check the obtained equation by plugging in other points along the tangent line and ensuring they satisfy the equation.

[tex]\hrulefill[/tex]

Step 1:

[tex](x_0,y_0) \rightarrow (-4,9)[/tex]

Step 2:

[tex]f(x)=6+\sqrt{5-x}\\\\\\\Longrightarrow f(x)=6+(5-x)^{1/2}\\\\\\\Longrightarrow f'(x)=\dfrac{1}{2} (5-x)^{1/2-1} \cdot -1\\\\\\\therefore \boxed{f'(x)=-\dfrac{1}{2\sqrt{5-x} } }[/tex]

Step 3:

[tex]f'(x)=-\dfrac{1}{2\sqrt{5-x} } ; \ (-4,9)\\\\\\\Longrightarrow f'(-4)=-\dfrac{1}{2\sqrt{5-(-4)} } \\\\\\\Longrightarrow f'(-4)=-\dfrac{1}{2\sqrt{9}} \\\\\\\Longrightarrow f'(-4)=-\dfrac{1}{2(3)} \\\\\\\therefore \boxed{ f'(-4)=m=-\dfrac{1}{6} }[/tex]

Step 4 and 5:

[tex]y-y_0=m(x-x_0)\\\\\\\Longrightarrow y-9=-\dfrac{1}{6}(x-(-4)) \\\\\\\Longrightarrow y-9=-\dfrac{1}{6}(x+4) \\\\\\\Longrightarrow y-9=-\dfrac{1}{6}x-\dfrac{2}{3}\\\\ \\\therefore \boxed{\boxed{ y=-\dfrac{1}{6}x-\dfrac{25}{3}}}[/tex]

Thus, the problem is solved.

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