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Vlada [557]
2 years ago
15

Mia, jade, and Amelia sell car wash tickets. Jade sells 8 more tickets then Mia. Amelia sells twice as many tickets as Mia. The

friends sell a total of 296 tickets. How many tickets does each friend sell?
Mathematics
1 answer:
Ede4ka [16]2 years ago
8 0

Answer:

Mia sold 72 tickets, Jade sold 80 tickets, and Amelia sold 144 tickets.

Step-by-step explanation:

You can solve this question by coming up with equations for each friend and then substituting them into the equation M + J + A = 296

Jade sells 8 more tickets than Mia, so Jade's equation for how many tickets she sold is the amount of tickets that Mia sold + 8:

J = M + 8

Amelia sells twice as many tickets as Mia, so her equation is the amount of tickets Mia sold * 2:

A = 2 * M

Now we can substitute these equations into the overall one.

M + (M + 8) + (2M) = 296

Now we can combine like terms and solve for M.

2M + 8 + 2M = 296

4M + 8 = 296

4M = 288

M = 72

We now have the amount of tickets Mia sold, so we can insert this number into the equations for Amelia and Jade.

J = M + 8

J = 72 + 8

J = 80

A = M * 2

A = 72 * 2

A = 144

So Mia sold 72 tickets, Jade sold 80 tickets, and Amelia sold 144 tickets.

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kodGreya [7K]

Answer:

(1) g[f(x)]=\frac{8x-1}{12x-4}

(2)  g^{-1}(x)=\frac{x+1}{(3x-2)}

(3)    x=\frac{9+\sqrt{129} }{6},  \frac{9-\sqrt{129} }{6}

Step-by-step explanation:

Given functions f(x) = (4x-1)

are                     g(x)=\frac{2x+1}{3x-1}

(1)  g[f(x)]=\frac{2(4x-1)+1}{3(4x-1)-1}

                =\frac{8x-2+1}{12x-3-1}

                =\frac{8x-1}{12x-4}

(2) for g^{-1}(x), rewrite the function g(x) in terms of an equation

y=\frac{2x+1}{3x-1}

Substitute y in place of x and x in place of y, then solve for y.

x=\frac{2y+1}{3y-1}

(3y-1)x = 2y+1

3xy - x = 2y + 1

3xy - 2y = x + 1

y(3x-2) = x + 1

y=(\frac{x+1}{3x-2} )

⇒ g^{-1}(x)=\frac{x+1}{(3x-2)}

(3)    f[g(x)]=(x-1)

       f[g(x)]=4[\frac{2x+1}{3x-1}] =(x-1)

⇒    \frac{(8x+4)-(3x-1)}{3x-1}=(x-1)

⇒    \frac{8x-3x+4+1}{3x-1}=(x-1)

⇒    \frac{5x+5}{(3x-1)}=(x-1)

⇒    5x+5=(3x-1)(x-1)

⇒    5x+5=3x^2-3x-x+1

⇒   5x+5=3x^2-4x+1

⇒   3x^2-9x-4=0

⇒    x=\frac{9\pm\sqrt{(-9)^2-4(-4)\times3} }{2\times3}

   x=\frac{9\pm\sqrt{81+48} }{6}

   x=\frac{9\pm\sqrt{129} }{6}

   x=\frac{9+\sqrt{129} }{6},  \frac{9-\sqrt{129} }{6}

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Answer:

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Step-by-step explanation:

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