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Varvara68 [4.7K]
3 years ago
8

Bob and Marie like to go to a good steakhouse. Dinners are $14 to $16 dollars with soup or salad. A beverage is $1.25. Dessert i

s $4.25. If one dinner is $14, the other is $16 and each diner has a beverage and dessert, what will the cost be if a 7% tax is charged and a 15% tip on the pre-tax amount is left? $43.32 $50.00 $50.02 $50.45
Mathematics
2 answers:
rosijanka [135]3 years ago
7 0

Answer:

50.02

Step-by-step explanation:

Bad White [126]3 years ago
3 0

Answer:

$50.02

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

See the attached figure.

For a quadrilateral inscribed in a circle

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So, ∠D + ∠B = 180°

From the figure:

∠B = x°  and ∠D = (2x+16)°

∴  <u>(2x + 16)° + (x)° = 180°</u>

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Delicious77 [7]

Answer:

Part 5.1.1:

\displaystyle \cos 2A = \frac{7}{8}

Part 5.1.2:

\displaystyle \cos A = \frac{\sqrt{15}}{4}

Step-by-step explanation:

We are given that:

\displaystyle \sin 2A = \frac{\sqrt{15}}{8}

Part 5.1.1

Recall that:

\displaystyle \sin^2 \theta + \cos^2 \theta = 1

Let θ = 2<em>A</em>. Hence:

\displaystyle \sin ^2 2A + \cos ^2 2A = 1

Square the original equation:

\displaystyle \sin^2 2A = \frac{15}{64}

Hence:

\displaystyle \left(\frac{15}{64}\right) + \cos ^2 2A = 1

Subtract:

\displaystyle \cos ^2 2A = \frac{49}{64}

Take the square root of both sides:

\displaystyle \cos 2A = \pm\sqrt{\frac{49}{64}}

Since 0° ≤ 2<em>A</em> ≤ 90°, cos(2<em>A</em>) must be positive. Hence:

\displaystyle \cos 2A = \frac{7}{8}

Part 5.1.2

Recall that:

\displaystyle \begin{aligned}  \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &=   1- 2\sin^2\theta \\ &= 2\cos^2\theta - 1\end{aligned}

We can use the third form. Substitute:

\displaystyle \left(\frac{7}{8}\right) = 2\cos^2 A - 1

Solve for cosine:

\displaystyle \begin{aligned} \frac{15}{8} &= 2\cos^2 A\\ \\ \cos^2 A &= \frac{15}{16} \\ \\ \cos A& = \pm\sqrt{\frac{15}{16}} \\ \\ \Rightarrow \cos A &= \frac{\sqrt{15}}{4}\end{aligned}

In conclusion:

\displaystyle \cos A = \frac{\sqrt{15}}{4}

(Note that since 0° ≤ 2<em>A</em> ≤ 90°, 0° ≤ <em>A</em> ≤ 45°. Hence, cos(<em>A</em>) must be positive.)

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Answer:

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Step-by-step explanation:

2w +  \frac{1}{2} ( - 40w + 12) \\ 2w - 20w + 6 \\  - 18w + 6 \\

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