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marusya05 [52]
3 years ago
7

a comedy show sells lower level tickets for $78 and upper level tickets for $59. on the opening knight 2368 tickets were sold fo

r total and $157800
Mathematics
1 answer:
shusha [124]3 years ago
3 0
Let L represent lower level tickets and U upper level.

Starter equations:
L + U = 2368
78L + 59U = 157800

Solve simultaneous equations:
59L + 59U = 139712
so 19L = 18088
L = 952
U = 2368 - 952 = 1416

952 lower level tickets and 1416 upper level tickets were sold.
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Find the local and global extrema for the graph of ƒ(x) = x(25 – x).
Allisa [31]

fx=25x-x²

dy/dx=0

25-2X=0

(=)X=14,5

4 0
3 years ago
Convert 68°C to degrees Fahrenheit. If necessary, round your answer to the nearest tenth of a degree
Karolina [17]

Answer: 154.4 Fahrenheit

Step-by-step explanation: 68°C = ( 68× 9 / 5 + 32 ) = 154.4°F

6 0
3 years ago
Use a calculator to find a decimal approximation for the following trigonometric function sin 28°48
Bumek [7]

The decimal approximation for the trigonometric function sin 28°48' is

Given the trigonometric function is sin 28°48'

The ratio between the adjacent side and the hypotenuse is called cos(θ), whereas the ratio between the opposite side and the hypotenuse is called sin(θ). The sin(θ) and cos(θ) values for a given triangle are constant regardless of the triangle's size.

To solve this, we are going to convert 28°48' into degrees first, using the conversion factor 1' = 1/60°

sin (28°48') = sin(28° ₊ (48 × 1/60)°)

= sin(28° ₊ (48 /60)°)

= sin(28° ₊ 4°/5)

= sin(28° ₊ 0.8°)

= sin(28.8°)

= 0.481753

Therefore sin (28°48') is 0.481753.

Learn more about Trigonometric functions here:

brainly.com/question/25618616

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5 0
2 years ago
Two altitudes of a triangle have lengths $12$ and $14$. What is the longest possible integer length of the third altitude
nataly862011 [7]

The longest possible altitude of the third altitude (if it is a positive integer) is 83.

According to statement

Let h is the length of third altitude

Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.

From Area of triangle

A = 1/2*B*H

Substitute the values in it

A = 1/2*a*12

a = 2A / 12 -(1)

Then

A = 1/2*b*14

b = 2A / 14 -(2)

Then

A = 1/2*c*h

c = 2A / h -(3)

Now, we will use the triangle inequalities:

  • a < b+c

2A/12 < 2A/14 + 2A/h

Solve it and get

h<84

  • b < a+c

2A/14 < 2A/12 + 2A/h

Solve it and get

h > -84

  • c < a+b

2A/h < 2A/12 + 2A/14

Solve it and get

h > 6.46

From all the three inequalities we get:

6.46<h<84

So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.

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8 0
2 years ago
Please help!<br> thankjs
olga_2 [115]

Answer:

20th: 100 nth: 5

Step-by-step explanation:

the table is multiplying by 5

so 5 × 20 = 100

8 0
3 years ago
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