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Galina-37 [17]
4 years ago
5

Beach hotel in Cancun is offering two weekend special.one includes a 2 night stay with 3 meals and cost $195. The other includes

a 3 night stay with 5 meals and cost $300. What is the cost of a single meals
Mathematics
2 answers:
daser333 [38]4 years ago
8 0
It meal cost 20 Because its 300 divided by 15
PSYCHO15rus [73]4 years ago
5 0
$75 per night and $15 per meal
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Vicky drew a scale drawing of a city. She used the scale 1 inch : 2 yards. If the actual width of a neighborhood park is 62 yard
klemol [59]

Answer:

The park is <u>31 inches</u> wide in the drawing.

Step-by-step explanation:

Given:

Vicky drew a scale drawing of a city. She used the scale 1 inch : 2 yards.

The actual width of a neighborhood park is 62 yards.

Now, to find the width of park in drawing.

Let the width of park in drawing be x.

The scale drawing of the city is 1 inch : 2 yards.

So, 1 inch is equivalent to 2 yards.

Thus, x is equivalent to 62 yards.

Now, to get the width of park in drawing by using cross multiplication method:

\frac{1}{2} =\frac{x}{62}

By cross multiplying we get:

62=2x

Dividing both sides by 2 we get:

31=x

x=31\ inches.

Therefore, the park is 31 inches wide in the drawing.

8 0
3 years ago
Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell
Tanzania [10]

Answer:

Length (parallel to the x-axis): 2 \sqrt{2};

Height (parallel to the y-axis): 4\sqrt{2}.

Step-by-step explanation:

Let the top-right vertice of this rectangle (x,y). x, y >0. The opposite vertice will be at (-x, -y). The length the rectangle will be 2x while its height will be 2y.

Function that needs to be maximized: f(x, y) = (2x)(2y) = 4xy.

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: g(x, y) =k. In this case

\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}.

Start by finding the first derivatives of f(x, y) and g(x, y)with respect to x and y, respectively:

  • f_x = y,
  • f_y = x.
  • \displaystyle g_x = \frac{x}{2},
  • \displaystyle g_y = \frac{y}{8}.

This method asks for a non-zero constant, \lambda, to satisfy the equations:

f_x = \lambda g_x, and

f_y = \lambda g_y.

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

g(x, y) = k.

\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right..

Replace the y in equation (b) with the right-hand side of equation (b).

\displaystyle x = \lambda \frac{\lambda \cdot \dfrac{x}{2}}{8} = \frac{\lambda^{2} x}{16}.

Before dividing both sides by x, make sure whether x = 0.

If x = 0, the area of the rectangle will equal to zero. That's likely not a solution.

If x \neq 0, divide both sides by x, \lambda = \pm 4. Hence by equation (b), y = 2x. Replace the y in equation (c) with this expression to obtain (given that x, y >0) x = \sqrt{2}. Hence y = 2x = 2\sqrt{2}. The length of the rectangle will be 2x = 2\sqrt{2} while the height will be 2y = 4\sqrt{2}. If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.

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3 years ago
PLEASE HELP! BRAINLIEST PLUS 50 POINTS! AND DO NOT GUESS! TYSM!
aivan3 [116]
Well i graph each one of those and the first choice seems to be correct. 
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3 years ago
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HELP What is the difference between Charlotte's lap time for Trial 1 and the average lap time for the track? 36.9 - 34.5 =​
Nookie1986 [14]
I feel Iike the answer but I kinda don’t understand it 2.4
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3 years ago
Given a can with a volume of 42 and a diameter of 4, what is the volume of a cone that fits perfectly inside the can? (Hint: onl
V125BC [204]

Answer:

              \large\boxed{\large\boxed{14 units^3}}

Explanation:

The<em> cone</em> is said to <em>fit perfectly inside the can</em> when the diameter of the cone and its height are equal to those of the can.

The shape of a can can be considered a cylinder.

The<em> volume of a cone that fits into the can</em> is 1/3 the volume of the can.

Thus, the volume of our cone is (1/3) × 42 unit³ = 14 units³

8 0
4 years ago
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