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3 years ago

6

For a rollerblading party there are 50 tickets available. Jason buys 6 tickets and Heather buys three times as many as Jason bou

ght. How many tickets are left to be sold?

2 answers:

garik1379 [7]3 years ago

8 0

There should be 26 tickets left.

Dmitry [639]3 years ago

5 0

There are 26 tickets left to be sold.

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Jimmy's school is selling tickets to the

natka813 [3]

Children’s tickets are $4, adult tickets are $14

Explanation:

4a+6c=80

If they made $4 more dollars on the second day than the first by selling one extra child ticket then we know a child’s ticket is $4 per ticket.

4a+6*4=80
4a+24=80
4a=56
14=a

So each adult ticket costs $14.

You can check by filling in $14 and $4 for each equation.

4*14+6*4=80

4*14+7*4=84

3 0

3 years ago

Which property is demonstrated <br> 9 * 3 * 4 = 4 * 3 * 9

mr_godi [17]

The commutative property of multiplication.

4 0

3 years ago

I NEED HELP PEOPLE!!!! WILL GIVE BRAINLIEST!!!!

Sauron [17]

Answer:

L(2,4) M(3,1) N(0,4)

Step-by-step explanation:

6 0

4 years ago

A light bulb consumes 3120 watt-hours in 3 days and 6 hours. How many watt-hours does it consume per day?

matrenka [14]

2850*24/(4*24+18)

The light bulb consumes 600 watt-hours per day.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=9%5C%3A%20%20%5C%3A%20%20%5Cfrac%7B%20%5Csin%28%20%5Ctheta%29%20%7D%7B1%20%2B%20%20%5Ccos%28%2

PIT_PIT [208]

Option (b) is your correct answer.

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{sin\theta }{1 + cos\theta } \times \dfrac{1 - cos\theta }{1 - cos\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}$

So, using this, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{1 - {cos}^{2}\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this identity, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{{sin}^{2}\theta }$

$\rm \: = \: \dfrac{1 - cos\theta }{sin\theta }$

<u>Hence, </u>

$\\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta } = \: \dfrac{1 - cos\theta }{sin\theta } }}} \\$

3 0

2 years ago

Read 2 more answers