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Serhud [2]
4 years ago
7

A total of 510 were sold for the school play. they were either adult tickets or student tickets. there were 60 more student tick

ets sold than adult tickets .How many adult tickets were sold?
Mathematics
1 answer:
mote1985 [20]4 years ago
4 0
I think it would be 450 because 510 - 60 = 450

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How much money does Mr.Williams need to budget for the museum tickets and parking?
Scrat [10]
I pretty sure the answer is $1086.80
okay so first you have the figure out how many students and teachers they're are gonna be. 
so if there's 171 students, and you need 2 adults for every 9 students then you need to divide 171 by 9 because it's 9 to a group so 1721 divided by 9 is 19. now you need to divide that by 2 because theres 19 groups of 9 and you need 2 adults for every 9, so 19 times 2 is 38. 
now that you have how many students and teachers are going you need to figure out how much is the total amount of tickets for everyone. if its 4.50 per student and 7.25 per teacher then you need to times $4.50 x 171 students and $7.25 x 38 adults. if you entered it right you should've gotten, $769.50 for the students and $275.50 for adults. add them both up. you should've gotten $1,045. 
okay so thats done with now the bus price. so if the bus price is 4% of the total price then you have to times .04 (4% in decimal) times $1045. the answer should be $41.80. so you $41.80 is the bus price now lastly you add that to the mount of the tickets. so last stop. $1045 + $41.80 = 1086.80 

So he needs to budget for the amount of $1086.80


I put the work down for you

4 0
3 years ago
Simplify the expression Write your answer as a power.
BaLLatris [955]
The answer is c to the fourth power.
7 0
3 years ago
Prove that the cube root of 2 is irrational.
harina [27]
The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, thatis, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must <span>be false, and the cube root of 2 is irrational.
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8 0
4 years ago
Find x. Round your answer to the nearest tenth of a degree. plz help!
Nuetrik [128]

Answer:

  36.9°

Step-by-step explanation:

The relevant trig relation is ...

  Cos = Adjacent/Hypotenuse

  cos(x) = 8/10 = 0.8

The inverse trig function is used to find the angle:

  x = arccos(0.8) ≈ 36.9°

7 0
3 years ago
Find the value of x and y
Serhud [2]
1) 180° - 146° = 34
    9х - 2 = 34
    9х = 34 + 2
    9х = 36
    х = 36 : 9
    х = 4
-------------

2) (9х - 2) + (6у + 44) = 180°
    (9 * 4 - 2) + (6у + 44) = 180
    (36 - 2) + (6у + 44) = 180
    34 + 6у + 44 = 180
    6у = 180 - 34 - 44
    6у = 102
    у = 102 : 6
    у = 17 
---------------
7 0
3 years ago
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