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slava [35]
3 years ago
6

The school that Kathryn goes to is selling tickets to a spring musical. On the first day

Mathematics
1 answer:
Nitella [24]3 years ago
3 0

Answer:

Senior citizen ticket = $4, student ticket = $6

Step-by-step explanation:

To do this, form a system of equations where x = the cost for a senior citizen ticket and y = the cost for a student ticket.

Since 12 senior citizen tickets and 14 students tickets cost $132, you get

12x + 14y = 132.

Since 12 senior citizen tickets and 10 students tickets cost $108, you get

12x + 10y = 108.

You can subtract those two equations, so

12x + 14y = 132

-(12x + 10y = 108)

and you get

4y = 24

divide both sides by 4

y = 6

substituting into the equation 12x + 10y = 108,

12x + 10(6) = 108

12x + 60 = 108

subtract 60 from both sides

12x = 48

divide both sides by 12

x = 4

A senior citizen ticket costs $4 and a student ticket costs $6.

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What are the solutions of 3(x – 4)(2x - 3) = 0? Check all that apply. -4 -3 win w bo 3 m/N 3 4​
Fudgin [204]

Answer:

Option (4) and Option (6)

Step-by-step explanation:

By the zero product property,

If (x - a)(x - b) = 0

Then, (x - a) = 0 ⇒ x = a

(x - b) = 0 ⇒ x = b

By following the zero product property,

From the given equation,

3(x - 4)(2x - 3) = 0

x - 4 = 0 ⇒ x = 4

2x - 3 = 0 ⇒ x = \frac{3}{2}

Therefore, options (4) and (6) will be the correct options.

7 0
3 years ago
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Which rule explains why these triangles are congruent?
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GenaCL600 [577]

Answer:

12

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3 0
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How do you Simplify the expression completely.<br><br> c(4b3+3a)+b(2cb2−6ac) <br> Click
Dafna1 [17]

Answer:

c×3 (2 b^3 - 2 a b + a)

Step-by-step explanation:

Simplify the following:

c (4 b^3 + 3 a) + b (2 c b^2 - 6 a c)

Hint: | Factor common terms out of 2 c b^2 - 6 a c.

Factor 2 c out of 2 c b^2 - 6 a c:

c (4 b^3 + 3 a) + b×2 c (b^2 - 3 a)

Hint: | Pull a common factor out of c (4 b^3 + 3 a) + b×2 c (b^2 - 3 a).

Factor c out of c (4 b^3 + 3 a) + b×2 c (b^2 - 3 a), resulting in c ((4 b^3 + 3 a) + b×2 (b^2 - 3 a)):

c (4 b^3 + 3 a + 2 b (b^2 - 3 a))

Hint: | Distribute 2 b over b^2 - 3 a.

2 b (b^2 - 3 a) = 2 b^3 - 6 a b:

c (4 b^3 + 3 a + 2 b^3 - 6 a b)

Hint: | Group like terms in 4 b^3 + 3 a - 6 a b + 2 b^3.

Grouping like terms, 4 b^3 + 3 a - 6 a b + 2 b^3 = (4 b^3 + 2 b^3) - 6 a b + 3 a:

c (4 b^3 + 2 b^3) - 6 a b + 3 a

Hint: | Add like terms in 4 b^3 + 2 b^3.

4 b^3 + 2 b^3 = 6 b^3:

c (6 b^3 - 6 a b + 3 a)

Hint: | Factor out the greatest common divisor of the coefficients of 6 b^3 - 6 a b + 3 a.

Factor 3 out of 6 b^3 - 6 a b + 3 a:

Answer: c×3 (2 b^3 - 2 a b + a)

7 0
3 years ago
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