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jenyasd209 [6]
3 years ago
8

a stadium has 10508 VIP boxes the stadium is divided into 12 equal section 2 Premium sections on 10th and second a seat at the p

remium session cost 48 per game at the Saunders section cost 27 per game how many seats are there in each section ​
Mathematics
1 answer:
oee [108]3 years ago
8 0
Answer:
1). 875 seats
2). 25 rows in each section
3). $8400
4). Saving of $360
5). 2105 tickets remained unsold
6). x = 16
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What type of triangle has side lengths of 4 7 and 9
Reika [66]
Scalene triangle has all different lengths
3 0
3 years ago
Scott viewed 72 classic cars at a car show. Of the cars he viewed, 3/8 were sports cars. How many sports cars did Scott view?
katovenus [111]
Scott viewed 27 sports cars because:
1/8 of 72 is 9
And 9 multiplied by 3 would make that 3/8
Also, 9 multiplied by 3 is 27
So that's the answer. Boom.
4 0
3 years ago
A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
Mariulka [41]

Answer:

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

                           S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- Now use the surface area formula:

                           S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
4 years ago
Read 2 more answers
The area of a rectangle is 252 cm², and the length is 3 less than 2 times the width. Find the dimensions of the rectangle.
SSSSS [86.1K]

Answer:

21cm, 12cm

Step-by-step explanation:

width=x cm

length=2x-3 cm

area= 252cm²

l*b= 252 cm²

(2x-3)(x)=252

2x²-3x= 252

this is how I did it

we need to find the nearest square of the number (it should be lesser than the area)

that is 225, whose square root is 15.

since there is a coefficient to the x², the value of x would be calculate like this 15- (2+1)= 15-3 = 12

x= 12

we can verify it too,

2*(12)² - 3*12=252

288-36=252

252=252

therefore width is 12 cm and width is 2*12-3= 24-3= 21 cm

verify;

21*12= 252

252=252

hence the dimensions are 12 am (width) and 21 cm(length).

3 0
3 years ago
Does the following represent the Transitive Property?
Sauron [17]
No it does not. If q=b and b=t then q=t is the correct answer. 
7 0
3 years ago
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