Ask question

Ask question

All categories

3 years ago

12

You and some friends are going to the fair. Small rides costs 4 tickets and while big rides costs 7 tickets. You and your friend

s bought a total of 250 tickets. Write an inequality to represent the different combinations of small and big you can enter.

1 answer:

tensa zangetsu [6.8K]3 years ago

7 0

Answer:

<em>4x+7y≤250</em>

Step-by-step explanation:

<u>Inequalities</u>

Let's call:

x = number of small rides

y = number of big rides

Since each small ride costs 4 tickets, x rides cost 4x tickets.

Since each big ride costs 7 tickets, y rides cost 7y tickets.

The total number of tickets is the sum of both:

Total tickets = 4x+7y

The friends have 250 tickets, so the total tickets of the rides cannot be greater than the 250 tickets available, thus:

4x+7y≤250

You might be interested in

Given this unit circle, what is the value of x? (x, -7/10)(1,0)

Paraphin [41]

Answer:

x = - $\frac{\sqrt{51} }{10}$

Step-by-step explanation:

the equation of a circle centred at the origin is

x² + y² = r² ( r is the radius )

The radius of a unit circle is r = 1

substitute (x, - $\frac{7}{10}$ ) into the equation and solve for x

x² + (- $\frac{7}{10}$ )² = 1²

x² + $\frac{49}{100}$ = 1 ( subtract $\frac{49}{100}$ from both sides )

x² = 1 - $\frac{49}{100}$ = $\frac{51}{100}$ ( take square root of both sides )

x = ± $\sqrt{\frac{51}{100} }$ = ± $\frac{\sqrt{51} }{10}$

since the point is in the 3rd quadrant then x < 0

x = - $\frac{\sqrt{51} }{10}$

5 0

2 years ago

For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.

andriy [413]

Answer:

If $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

The signal of $\bigtriangleup$ determines how many real roots an equation has:

$\bigtriangleup > 0$ : Two real and different solutions

$\bigtriangleup = 0$ : One real solution

$\bigtriangleup < 0$ : No real solutions

In this problem, we have the following second order polynomial:

$(k+1)x^{2} + 4kx + 2 = 0$ .

This means that $a = k+1; b = 4k; c = 2$

It has one solution if

$\bigtriangleup = 0$

$b^{2} - 4ac = 0$

$16k^{2} -8(k+1) = 0$

$16k^{2} - 8k - 8 = 0$

We can simplify by 8

$2k^{2} - k - 1 = 0$

The solution is:

$k = 1$ or $k = -\frac{1}{2}$

So, if $k = 1$ or $k = -\frac{1}{2}$ , there is only one solution to the given quadratic equation.

4 0

3 years ago

Simplify: m^2n^3 aanm^2a^2n^4

dolphi86 [110]

9514 1404 393

Answer:

a^4·m^4·n^8

Step-by-step explanation:

The applicable rule of exponents is ...

(a^b)(a^c) = a^(b+c)

__

m^2n^3 aanm^2a^2n^4

= a^(1+1+2)·m^(2+2)·n^(3+1+4)

= a^4·m^4·n^8

3 0

3 years ago

How can i find the surface area of a rectangle

omeli [17]

Answer:

Legenth times width

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Plz help me I can't get these wrong plz help meeeee

Vika [28.1K]

Answer: w<0

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers