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dimaraw [331]
3 years ago
12

Tickets for a minor league baseball game for an adult and two children cost a total of $21 the adult ticket is three dollars mor

e than a child's ticket. Find the cost of the child's ticket and an adult ticket.
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
6 0

Answer:the cost of one child's ticket is $6

the cost of one adult ticket is $9

Step-by-step explanation:

Let x represent the cost of one child's ticket

Let y represent the cost of one adult ticket.

Tickets for a minor league baseball game for an adult and two children cost a total of $21. This means that

2x + y = 21 - - - - - - - - - - 1

The adult ticket is three dollars more than a child's ticket. This means that

y = x + 3

Substituting y = x + 3 into equation 1, it becomes

2x + x + 3 = 21

3x + 3 = 21

3x = 21 - 3 = 18

x = 18/3 = 6

y = x + 3 = 6 + 3

y = 9

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Answer:

Solving the equation x^2 - 6x + 10 = 0 we get x=3+i \ , \ x=3-i

Step-by-step explanation:

Noah's step to solve the equation x^2 - 6x + 10 = 0 are:

x^2 - 6x + 10 = 0\\x^2 - 6x = -10\\x^2 - 6x + 9 = -10 + 9\\(x-3)^2 = -1

The next step is to take square root on both sides because: \sqrt{x^2}=x

\sqrt{(x-3)^2}=\sqrt{-1}  \\x-3=\pm\sqrt{-1} \\x=\pm(\sqrt{-1})+3 \\

Now, we know that: \sqrt{-1}=i

x=\pm(i)+3\\x=3+i \ , \ x=3-i

So, Solving the equation x^2 - 6x + 10 = 0 we get x=3+i \ , \ x=3-i

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3 years ago
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Grant plans to evaporate enough water from 22 gallons of a 16% ammonia solution to make a 24% ammonia solution. Which equation c
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Jim is 4 ft from a lamppost that is 16 ft tall and he is 6 foot tall his shadow is ____?____ ft long?
Stolb23 [73]

 

To solve this, let us first imagine a smaller triangle created by the head of Jim (A), the top of the lamp post (B), and somewhere on the body of the lamp post which is directly perpendicular to the head of Jim (C).

 

CB = 16 – 6 = 10 ft

AC = 4 ft

 

Calculate for angle B using tan function:

tan B = AC / CB

B = tan^-1 (4 / 10)

B = 21.8°

 

Now imagine a bigger triangle created by the tip of shadow (D), the top of the lamp post (B), and the base of the lamp post (E).

BE = 16 ft

B = 21.8°

 

We can calculate for DE using tan function:

tan B = DE / BE

(16 ft) tan 21.8 = DE

DE = 6.4 ft

 

Since Jim is 4ft away from the base of the lamp post, therefore the length of the shadow is:

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Therefore the length of Jim’s shadow is 2.4 ft long

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3 years ago
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