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Serga [27]
3 years ago
15

Part II: Find the sine, cosine, and tangent ratios of <y ​

Mathematics
1 answer:
Lorico [155]3 years ago
6 0

Answer:

sin θ = 22.62°

cos θ = 22.62°

tan θ = 22.62°

Step-by-step explanation:

From the diagram above

Opposite = 5

Adjacent = 12

Hypotenuse = 13

a) sin θ = Opp/Hypotenuse

sin θ = 5/13

sin θ = 0.3846153846

arc sin 0.3846153846

= 22.619864948°

Approximately = 22.62°

b) cos θ = Adjacent/ Hypotenuse

θ = 12/13

= arccos(0.9230769230769231)

22.619864948°

Approximately = 22.62°

c) tan θ = Opposite/Adjacent

θ = 5/12

= arctan(0.4166666666666667)

= 22.619864948°

= 22.62°

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If 2y^2+2=x^2, then find d^2y/dx^2 at the point (-2, -1) in simplest form.​
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Answer:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}

Step-by-step explanation:

We have the equation:

2y^2+2=x^2

And we want to find d²y/dx² at the point (-2, -1).

So, let's take the derivative of both sides with respect to x:

\frac{d}{dx}[2y^2+2]=\frac{d}{dx}[x^2]

On the left, let's implicitly differentiate:

4y\frac{dy}{dx}=\frac{d}{dx}[x^2]

Differentiate normally on the left:

4y\frac{dy}{dx}=2x

Solve for the first derivative. Divide both sides by 4y:

\frac{dy}{dx}=\frac{x}{2y}

Now, let's take the derivative of both sides again:

\frac{d}{dx}[\frac{dy}{dx}]=\frac{d}{dx}[\frac{x}{2y}]

We will need to use the quotient rule:

\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}

So:

\frac{d^2y}{dx^2}=\frac{\frac{d}{dx}[(x)](2y)-x\frac{d}{dx}[(2y)]}{(2y)^2}

Differentiate:

\frac{d^2y}{dx^2}=\frac{(1)(2y)-x(2\frac{dy}{dx})}{4y^2}

Simplify:

\frac{d^2y}{dx^2}=\frac{2y-2x\frac{dy}{dx}}{4y^2}

Substitute x/2y for dy/dx. This yields:

\frac{d^2y}{dx^2}=\frac{2y-2x\frac{x}{2y}}{4y^2}

Simplify:

\frac{d^2y}{dx^2}=\frac{2y-\frac{2x^2}{2y}}{4y^2}

Simplify. Multiply both the numerator and denominator by 2y. So:

\frac{d^2y}{dx^2}=\frac{4y^2-2x^2}{8y^3}

Reduce. Therefore, our second derivative is:

\frac{d^2y}{dx^2}=\frac{2y^2-x^2}{4y^3}

We want to find the second derivative at the point (-2, -1).

So, let's substitute -2 for x and -1 for y. This yields:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(-1)^2-(-2)^2}{4(-1)^3}

Evaluate:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2(1)-(4)}{4(-1)}

Multiply:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{2-4}{-4}

Subtract:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{-2}{-4}

Reduce. So, our answer is:

\frac{d^2y}{dx^2}_{(-2, -1)}=\frac{1}{2}

And we're done!

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The graph illustrated is divided into three seperate sections namely <em>A, B and C,</em> the short story culled from the graph is described thus:

Rate of change or slope = (Rise/ Run)

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Distance traveled = 50 meters

<u>SECTION B</u> :

This section gives an horizontal line, hence, the slope is 0.

Distance covered = 0 ; as there was no change in the distance over the period of time.

<u>SECTION</u><u> </u><u>C</u><u>:</u>

Slope = (80 - 50) ÷ (70 - 60) = 30/10 = 3

The total distance traveled = 80 - 50 = 30 meters

<u>The total </u><u>distance</u><u> </u><u>and</u><u> </u><u>time spent traveling</u> :

Total distance traveled = (50 + 30) = 80 meters

Total time spent = 70 seconds (<em>time at the 80 meter mark</em>)

Hence, the story.

Learn more : brainly.com/question/25309609

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