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Tema [17]
4 years ago
14

If a stadium has 15,000 seats sold at $10.00, $12.50, and $15.00 equally distributed in three sections, how much can be made if

the stadium sells out?
Mathematics
1 answer:
MissTica4 years ago
8 0
First we have to determine how many people fir in each section. 
Since these sections are evenly distributed, each section has a third of the capacity, thus= 15,000/3 = 5000 persons per section.

Now we determine revenue by section by multiplying times the ticket price.

5,000 * $10.00 = $50,000
5,000 * $12.50 = $62,500
5,000 * $15.00 = $75,000

We add up the totals for $187,500 and that's the answer!
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Solve for c in the literal equation 8p +9c =p. Represent any fractions in reduced proper or improper
photoshop1234 [79]

Answer:

c = -7/9p

Step-by-step explanation:

8p + 9c = p

8p − 8p + 9c = p − 8p

Step 2:  

9c = −7p

Step 3:  

9c/9 = -7p/9

c = -7/9p

Hope it helps, I did the quiz

3 0
3 years ago
I'm unsure how to solve for w, can someone please help?
iren2701 [21]

Answer:

w = 1/2

Step-by-step explanation:

One of the ways to solve problems using log is to use these two equations that use the same variables and plug in numbers.

log b (a) = c

b^c = a

By plugging in the numbers we get this:

log 16 (4) = w

16^w = 4

w = 1/2

4 0
3 years ago
A dump truck is unloading sand that falls into a conical pile. Find the height of the cone when the diameter of the pile is 16 f
Rudiy27

Answer:

8.21

Step-by-step explanation:

formula is:

V= 3\frac{volume}{pie(diameter)}

you plug everything in but when you go to plug in your diameter you need to use the radius. the radius is always half of the diameter.

6 0
3 years ago
La expresion 18+2^3 / 4x2
erik [133]
18+8/4x2
18+8/8
18+1
PEMAS
parentheses,
7 0
4 years ago
Please help me to prove this!​
Sophie [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C                → A = C - B

                                          → B = C - A

Use the Double Angle Identity:     cos 2A = 2 cos² A - 1

                                             → (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS:                     cos² A + cos² B + cos² C

\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C

\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C

\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C

\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]

\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)

\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)

\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C   \checkmark

5 0
3 years ago
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