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emmainna [20.7K]
1 year ago
14

A total of 750 tickets were sold for the school play. They were either adult tickets or student tickets. There were 50 more stud

ent tickets sold than adult tickets.
How many adult tickets were sold?
1 adult tickets
х
5
?
Mathematics
1 answer:
Vlada [557]1 year ago
5 0

Answer:

350 adult tickets were sold.

Step-by-step explanation:

Let's set up some values:

x = adult tickets

x + 50 = student tickets, since there were 50 more student tickets than adult tickets.

Now, let's set up our equation. If these two values add up to 750 tickets, then we can conclude the following:

(x) + (x + 50) = 750

2x + 50 = 750

2x = 700

x = 350 adult tickets. This means there were 400 student tickets sold.

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Answer:

IF I am correct, 5

Step-by-step explanation:

98/ 18

8 0
2 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
-1/19,7/4,-4/7 least to greatest
oee [108]
Answer:

Order from Least to Greatest
-4/7  <  -1/19  <  7/4


Showing Work
Rewriting as fractions or any negatives if necessary:
-1/19, 7/4, -4/7

The least common denominator (LCD) is: 532.

Rewriting as equivalent fractions with the LCD:
-28/532, 931/532, -304/532

Ordering these fractions by the numerator:
-304/532  <  -28/532  <  931/532

Therefore, the order of your input is:
-4/7  <  -1/19  <  7/4

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3 years ago
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See attached PDF. (The censor thinks there are some unseemly words in there.)

Download pdf
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1.What value goes in the blank to make this a proportion:
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1. 24
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