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2 years ago

13

Traveling carnivals move from town to town, staying for a limited number of days before moving to the next stop. The management

of a certain carnival knows that each
time it opens in a new town, it can expect to bring in about $15,000 in revenue the first night. Each night after the first revenue will be about 75% of the previous night's
revenue.
To the nearest dollar, about how much revenue can the carnival anticipate on its 5th night in town?

2 answers:

svetoff [14.1K]2 years ago

7 0

Are they answer choices?

zhuklara [117]2 years ago

6 0

Answer: 4746.09375 dollars (just round down to nine cents)

Step-by-step explanation:

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Elza [17]

Answer:

$y = 3x + 10$

$y = 3x + 15$

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2 years ago

6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the

MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

$a_1$ the horizontal distance between the first observer and the ballonist

$a_2$ the horizontal distance between the second observer and the ballonist

$\alpha _1$ and $\alpha _2$ the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

$S=a*h$ (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

$S=S_1+S_2$ (equation 2)

$S_1=a_1*h$

But we can write $S_1$ in terms of $\alpha _1$ , like this:

$\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}$

And for $S_2$ will be the same:

$S_2=\frac{h^{2} }{\tan(\alpha _2)}$

Replacing in the equation 2:

$S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})$

And replacing in the equation 1:

$h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}$

So, we can replace all the known data in the last equation:

$h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft$

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

6 0

2 years ago

The base of a solid right pyramid is a regular hexagon with

cupoosta [38]

Answer: Last option.

Step-by-step explanation:

Observe the base of the pyramid, which is a regular hexagon. You can see that there is a right triangle.

The area of the triangle can be calculated with this formula:

$A_t=\frac{bh}{2}$

Where "b" is the base and "h" is the height.

You can say that the apothem is the height of the right triangle, then, you need to find the base applying the Pythagorean Theorem:

$(2x)^2=(x\sqrt{3})^2+b^2\\\\b=\sqrt{(2x)^2-(x\sqrt{3})^2} \\\\b=\sqrt{4x^2-3x^2}\\\\b=\sqrt{x^2}\\\\b=x$

Then, the area of the triangle is:

$A_t=\frac{x(x\sqrt{3})}{2}=\frac{x^2\sqrt{3})}{2}$

Since the base of the pyramid is a regular hexagon, then you can multiply by 12 the area of the right triangle calculated above, in order to find the area of the hexagon. Then you get:

$A_h=12(\frac{x^2\sqrt{3})}{2})=6x^2\sqrt{3}\ units^2$

5 0

2 years ago

Read 2 more answers

A hackberry tree has roots that reach a depth of

Degger [83]

Answer:

24.6967 meters

Step-by-step explanation:

The roots of the tree go 6 and 5 over 12 meters below the ground level.

Now, 6 and 5 over 12 meters is equivalent to 6.4167 meters.

Again the top of the tree is 18.28 meters high from the ground level.

Therefore, the total height of the tree from the bottom of the root to the top is

(6.4167 +18.28) = 24.6967 meters (Answer)

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2 years ago

The equation y=−0.0088x2+0.79x+15 models the speed x (in miles per hour) and average gas mileage y (in miles per gallon) for a v

Neko [114]

we are given

$y=-0.0088x^2+0.79x+15$

where

y is average gas mileage in miles per gallon

x is speed in miles per hour

we have to find average gas mileage at a speed of 60 miles per hour

so, we have

$x=60mph$

now, we can plug it and find y

$y=-0.0088(60)^2+0.79(60)+15$

we get

$y=30.72$

so,

the best approximate for the average gas mileage at a speed of 60 miles per hour is 30.7 mi/gal...........Answer

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2 years ago

Read 2 more answers