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Zarrin [17]
3 years ago
15

A hotel charges $90 per night for a room. They charge an additional $8 for each person staying in the room. Each room has a maxi

mum capacity of 5 occupants. Write a function for this situation where p is the number of people in a room and c is the total cost. Is the function continuous or discrete?
Mathematics
1 answer:
Leya [2.2K]3 years ago
8 0

Your Answer Will Be 90+8p=c

P = How Many People Will Be Staying.

C = The Total Cost.

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Preston measured the button on his shirt. Then he calculated that it has an area of 50.24 square millimeters. What is the button
soldi70 [24.7K]
*note: for this problem you have to round a few values. I rounded to 2 decimal places, but your answer could be different if you have to round to more/less than 2 decimal places

Answer: Diameter = 8 millimeters


Buttons are circles, so we will use the formula for area of a circle:

A = πr^2

We are solving for the diameter, which is 2 times the radius:

d = 2r

So now we plug in what we know.
The area A = 50.24 square millimeters, therefore we plug it in and simplify:

50.24 square millimeters = πr^2

50.24/π = r^2

15.99 = r^2

squareroot(15.99) = r

r = 4 millimeters

Now we use the radius to calculate the diameter.

r = 4 millimeters

d = 2r

d = 2(4)

diameter = 8 millimeters
8 0
3 years ago
The light from a lamp creates a shadow on a wall with a hyperbolic border. Find the equation of the border if the distance betwe
dolphi86 [110]

This question is incomplete, the complete question is;

The light from a lamp creates a shadow on a wall with a hyperbolic border. Find the equation of the border if the distance between the vertices is 18 inches and the foci are 4 inches from the vertices. Assume the center of the hyperbola is at the origin.

Answer:

the equation of the border is x²/81 - y²/88 = 1

Step-by-step explanation:

Given the data in the question;

distance between vertices = 2a = 18 in

so

a = 18/2 = 9 in

distance of foci from vertices = 4 in

hence distance between two foci = 2c = 18 + 4 + 4 =  26

so

c = 26/2 = 13 in

Now, from Pythagorean theorem

b = √( c² - a² )

we substitute

b = √( (13)² - (9)² ) = √( 169 - 81 ) = √88

we know center is at the origin, so

( h, k ) = ( 0, 0 )

h = 0

k = 0

Using equation of hyperbola

[ ( x-h )² / a² ] - [ ( y - k )² / b² = 1

we substitute

[ ( x-0 )² / 9² ] - [ ( y - 0 )² / (√88)² ] = 1

x²/81 - y²/88 = 1

Therefore, the equation of the border is x²/81 - y²/88 = 1

3 0
3 years ago
Prove the identity 2csc2x=csc^2xtanz
Verizon [17]

Step-by-step explanation:

Consider the LHS, after the 5th step, consider the RHS

2 \csc(2x)  =  \csc {}^{2} (x)  \tan(x)

2 \frac{1}{ \sin(2x) }  =  \csc {}^{2} (x)  \tan(x)

2  \times \frac{1}{2 \sin(x)  \cos(x) }  =  \csc {}^{2} (x)  \tan(x)

\frac{1}{ \sin(x) \cos(x)  }  =   \csc {}^{2} (x)  \tan(x)

\csc(x)  \sec(x)  =  \csc {}^{2} (x)  \tan(x)

Consider the RHS

\csc(x)  \sec(x)  =( 1 +  \cot {}^{2} (x) ( \tan(x))

\csc(x)  \sec(x)  =  \tan(x)  +  \cot(x)

\csc(x)  \sec(x)  =  \frac{ \sin(x) }{ \cos(x) }  +  \frac{ \cos(x) }{ \sin(x) }

\csc(x)  \sec(x)  =  \frac{1}{ \sin(x) \cos(x)  }

\csc(x)  \sec(x)  =  \csc(x)  \sec(x)

7 0
2 years ago
Find the value of x.<br> 30°<br> 15x<br> A. 4<br> B. 10<br> C. 12<br> D. 2
goldenfox [79]

Answer:

10 is the answer

Step-by-step explanation:

5 0
3 years ago
Simplify the monomials
Tomtit [17]

Answer:

1. -12x^{5}y^{5}

2. \frac{10b^{8} }{a^{3} }

3. correct

4. a^{8}b^{10}c^{15}

Step-by-step explanation:

When dividing a monomial by another monomial, we divide the coefficients and apply the quotient law of exponents, x m ÷ x n = x m – n to the variables. If both the monomials have negative coefficients, the negative signs cancel out and the answer so obtained will be having a positive coefficient only.

For example, the binomial 2x +4 can be multiplied by the monomial 4x so that (2x) (4x) equals 8x2 and 4·4x equals 16x, using the distributive property. Therefore, (2x +4) (4x) equals 8x2 + 16x.Mar 15, 2016

Use the Quotient Property. When we divide monomials with more than one variable, we write one fraction for each variable. Use fraction multiplication. Simplify and use the Quotient Property.

7 0
3 years ago
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