The eatery restaurant has 200 tables on the reason even if there's no there's no reservation for one half of the tables were

The scale for the drawing of a rectangular playing field is 2 inches=7 feet. Find an equation you can use to find the dimensions

lina2011 [118]

Answer:

The actual answer is 7/2

Step-by-step explanation:

5 0

3 years ago

I HAVE UNTIL 3:30 PLZ HELPPPP 40 POINTS AND BRAINLIEST

stepan [7]

Answer:

you add 120 with x and get the answer and then you add the answer with 8 with y and then you add 12 to the answer and then you the real final answer

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Jennifer bought an umbrella on a business trip. The umbrella is 29 inches long.

OleMash [197]

Answer:

Answer 30^{2}

Step-by-step explanation:

24^{2} + 18^ {2} = c^{2}

576 + 324 = c^{2}

\sqrt{900} c^{2} = 30

hope this helps!

5 0

3 years ago

Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y =

Strike441 [17]

Answer:

The volume is:

$\displaystyle\frac{37\pi}{10}$

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

$\displaystyle\int_a^b\pi r^2dx$

Notice here the radius of the washer is the difference of the given curves:

$x-\sqrt{x}$

So the integral becomes:

$\displaystyle\int_1^4\pi(x-\sqrt{x})^2dx$

We solve it:

Factor $\pi$ out and distribute the exponent (you can use FOIL):

$\displaystyle\pi\int_1^4x^2-2x\sqrt{x}+x\,dx$

Notice: $x\sqrt{x}=x\cdot x^{1/2}=x^{3/2}$

So the integral becomes:

$\displaystyle\pi\int_1^4x^2-2x^{3/2}+x\,dx$

Then using the basic rule to evaluate the integral:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{2x^{5/2}}{5/2}+\frac{x^2}{2}\right|_1^4$

Simplifying a bit:

$\displaystyle\pi\left[\frac{x^3}{3}-\frac{4x^{5/2}}{5}+\frac{x^2}{2}\right|_1^4$

Then plugging the limits of the integral:

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(4)^{5/2}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Taking the root (rational exponents):

$\displaystyle\pi\left[\frac{4^3}{3}-\frac{4(2)^{5}}{5}+\frac{4^2}{2}-\left(\frac{1}{3}-\frac{4}{5}+\frac{1}{2}\right)\right]$

Then doing those arithmetic computations we get:

$\displaystyle\frac{37\pi}{10}$

6 0

3 years ago

Charlie is watching hot air balloons. balloon a has risen at a 50° angle. balloon b has risen at a 78° angle. if the distance fr

AfilCa [17]

The distance from Balloon A to Balloon B, d ≈ 1,052 feet

What is distance ?

Distance is defined to be the magnitude or size of displacement between two positions.
Note that the distance between two positions is not the same as the distance traveled between them.
Distance traveled is the total length of the path traveled between two positions. Distance traveled is not a vector.

The given parameters are;

The angle at which Balloon A rises, x° = 50° above horizontal

The angle at which Balloon B rises, y° = 78° above horizontal

The (vertical) distance of Balloon A to the ground, h = 1,000 ft.

The required parameter;

The distance from Balloon A to Balloon B, d

We find the distances of the balloons from Charlie and the angle between the balloon strings, θ, then apply cosine rule

Let,

The height of the balloon strings are equal

The distance of Balloon A to the ground, h₁ = The distance of Balloon B to the ground, h₂ = 1,000 ft.

The distance of the Balloon A from Charlie, l₁, is given as follows;

l₁ × sin(x°) = h₁

∴ l₁ = h₁/(sin(x°))

Which gives;

l₁ = 1,000 ft./(sin(50°)) ≈ 1,305.41 ft.

l₁ ≈ 1,305.41 ft.

For Balloon B, we get;

h₁ = h₂ = 1,000 ft.

∴ l₂ = 1,000 ft./(sin(78°)) ≈ 1,022.34 ft.

l₂ ≈ 1,022.34 ft

The angle between the balloon strings, θ = 180° - (x° + y°)

∴ θ = 180° - (50° + 78°) = 52°

The angle between the balloon strings, θ = 52°

By cosine rule, we have;

d = √(l₁² + l₂² - 2 × l₁ × l₂ × cos(θ))

∴ d = √(1,305.41² + 1,022.34² - 2 × 1,305.41×1,022.34 × cos(52°)) ≈ 1,052 feet

Therefore, the distance from Balloon A to Balloon B, d ≈ 1,052 feet

Learn more about distance

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7 0

2 years ago