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Nuetrik [128]
3 years ago
12

Pls pls help Show work :)

Mathematics
1 answer:
Natasha2012 [34]3 years ago
5 0

9514 1404 393

Answer:

  31.243 units

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.

  Sin = Opposite/Hypotenuse

  sin(30°) = BE/100

  BE = 100·sin(30°)

Then ...

  Tan = Opposite/Adjacent

  tan(58°) = BE/x

  x = BE/tan(58°) = 100·sin(30°)/tan(58°)

  x ≈ 31.243 . . . . units

_____

<em>Comment on the figure</em>

The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.

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Answer:

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Step-by-step explanation:

Area of circle= πr²

Substitute: π(5)²

Solution: 25π or about 78.5 cm²

7 0
3 years ago
Write an
Delicious77 [7]

Answer:

Slope=2.

Step-by-step explanation:

Slope=y1-y2/x1-x2

Where

X1,y1,=(-5, -11) X2,y2=(6,11).

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2 years ago
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Evaluate the interval (Calculus 2)
Darya [45]

Answer:

2 \tan (6x)+2 \sec (6x)+\text{C}

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x

\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}

If the terms are multiplied by constants, take them outside the integral:

\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x

Multiply by the conjugate of 1 - sin(6x) :

\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x

\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x

\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:

\implies \sin^2 (6x) + \cos^2 (6x)=1

\implies \cos^2 (6x)=1- \sin^2 (6x)

\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x

Expand:

\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x

\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:

\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x

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\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}

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\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}

Simplify:

\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}

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Learn more about indefinite integration here:

brainly.com/question/27805589

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3 0
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Let

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y----------> the total cost to rent a jet sky in dollars

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y=24.00+2.25x ---------> equation 1

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equate equation 1 and equation 2

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see the attached figure  

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Answer:

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= 31.92 - 24

= 7.92

4 0
3 years ago
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