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WINSTONCH [101]
2 years ago
10

2v^2-12 =-12v Would really appreciate it loves❤️

Mathematics
1 answer:
Black_prince [1.1K]2 years ago
7 0

For this case we have the following equation:

2v ^ 2-12 = -12v

Rewriting we have:

2v ^ 2 + 12v-12 = 0

Dividing by 2 to both sides of the equation:

v ^ 2 + 6v-6 = 0

We apply the quadratic formula:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

We have to:

a = 1\\b = 6\\c = -6

Substituting:

x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}

Thus, we have two roots:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

ANswer:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

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The cost of children’s ticket is $ 5

<h3><u>Solution:</u></h3>

Let "c" be the cost of one children ticket

Let "a" be the cost of one adult ticket

Given that adult ticket to a museum costs 3$ more than a children’s ticket

<em>Cost of one adult ticket = 3 + cost of one children ticket</em>

a = 3 + c  ------ eqn 1

<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>

200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100

200 \times a + 100 \times c = 2100

200a + 100c = 2100 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>

Substitute eqn 1 in eqn 2

200(3 + c) + 100c = 2100

600 + 200c + 100c = 2100

600 + 300c = 2100

300c = 1500

<h3>c = 5</h3>

Thus the cost of children’s ticket is $ 5

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