All categories

2 years ago

Adult tickets to a concert sold at $8, while student tickets cost $5.

Mathematics

2 answers:

geniusboy [140]2 years ago

4 0

Answer:

30 Tickets

Step-by-step explanation:

MrRissso [65]2 years ago

3 0

30 tickets

Step by step:
When you multiply 30x8 you get 240. You multiply by 8 because that is how much one adult ticket costs. And they tell you that they sold a total of 70 tickets, which means to find the number of children tickets that were sold would be 70-30=40. Then to prove that for sure, multiply...

40x5=200. You multiply by 5 because that is how much one children ticket costs.

Then when you add the totals of the adult tickets sold and the children tickets sold, you get $440, which means that 30 adult tickets were sold!!

You can use this same process for problems like these or to verify the answer!

Please give me brainliest!
Hope this helped!

You might be interested in

Item 7

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

2 years ago

the population of a country double every 10 years from 1950 to 1980.what was the percentage increase in population during this t

fredd [130]

700% increase.

Step-by-step explanation:

The duration was 3 decades, or 30 years. The population starts at 100%, then doubles to 200% (decade 1), then 400% (decade 2), then 800% (decade 3).

800% - 100% = 700%

3 0

3 years ago

If something is scarce it is generally

dmitriy555 [2]

Little of, not in plethora, small numbered

6 0

3 years ago

Please help!!!! Chris is a fitness instructor.

dmitriy555 [2]

Answer:

i think its d

Step-by-step explanation:

3 0

3 years ago

Ok guys my friend is telling me that i am doing this problem wrong 4^4+1/2=256.5 and he is tell me that it is 16.5 am i right or

Alexxx [7]

Simplify 4^4 to 256

256 + 1/2 = 256.5

Simplify 256 + 1/2 to 513/2

513/2 = 256.5

Since both sides equal, there are infinitely many solutions.

3 0

3 years ago

Read 2 more answers