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3 years ago

A total of

2 answers:

5 0

Let x be the number of adult tickets sold, such that 2x is the number of student tickets sold.

2x + x = 384, since the number of adult and student tickets add up to the total number of tickets.
3x = 384
x = 128

So, there are 128 adult tickets, leaving 256 student tickets sold.

densk [106]3 years ago

4 0

I believe it is 128. I may be wrong but if there are 384 tickets sold and there were 2 times more student tickets sold than adult then there are 128 tickets lift over to be adult tickets

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3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

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3 years ago

A ramp leads up to a building. The top of the ramp is 6 feet above the ground, and the bottom is 16 feet. What is the length of

andre [41]

We could use the Pythagorean theorem for this kind of problem I think:
A^2 + B^2 = C^2
6^2 + 16^2 = C^2
36 + 256 = C^2
292 = C^2
17.08 = C
C = 17.1 to the nearest tenth
Sorry if it’s wrong and glad I could help if it’s right ❤️❤️ Take care!

7 0

2 years ago