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].
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Solve (9th grade math)
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: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.
Differentiate the function. Then find an equation of the tangent line at the indicated point on the graph of the function.
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.
Find the original slope of (-6,-1) and (0,3)
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