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Monica [59]
3 years ago
11

URGENT, PLEASE ANSWER. Joshua was taking a trip to an amusement park with his family. To get his family in, his parents had to p

urchase two adult tickets, one child ticket, and one toddler ticket. The cost of an adult ticket was $45.50. A child ticket was 8/10 of the price of an adult ticket, and a toddler ticket was 2/5 of the price of an adult ticket. Part A: What was the cost of a child ticket? Part B: What was the cost of a toddler ticket?
Mathematics
1 answer:
ASHA 777 [7]3 years ago
5 0
Child ticket is $36.00
toddler ticket is $18.20
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3/8=0.375
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0.375<0.38<0.4

therefore: 3/8,0.38,2/5

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what is the perimeter measured in centimeters of the rectangle pictured below do not include units in your answer
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Answer:

183

Step-by-step explanation:

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So, the value of x is 1.19

The question is an exponential equation

<h3>What is an exponential equation?</h3>

An exponential equation is a mathematical expression between two quantities in which one variable is raised to a power of the other variable.

<h3>How to find x?</h3>

Since 4^{x} + 6^{x} = 9^{x},

Dividing through by 4ˣ, we have

\frac{4^{x} }{4^{x} } + \frac{6^{x} }{4^{x} }  = \frac{9^{x} }{4^{x} } \\1 + (\frac{6}{4})^{x} }  = (\frac{9}{4})^{x} } \\1 + (\frac{3}{2})^{x} }  = (\frac{3^{2} }{2^{2} })^{x} } \\1 + (\frac{3}{2})^{x} }  = (\frac{3}{2})^{2x} }

Let y = (3/2)ˣ

So,

1 + y = y²

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Using the quadratic formula to find y,

y = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}

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y = \frac{-(-1) +/- \sqrt{(-1)^{2} - 4\times 1 \times (-1)} }{2\times 1}\\= \frac{1 +/- \sqrt{1 + 4} }{2}\\= \frac{1 +/- \sqrt{5} }{2}\\= \frac{1 - \sqrt{5} }{2} or  \frac{1 + \sqrt{5} }{2}\\= \frac{1 - 2.236}{2} or  \frac{1 + 2.236}{2}\\= \frac{- 1.236}{2} or  \frac{3.236}{2}\\= -0.618 or 1.618

Since y = (3/2)ˣ

Takung logarithm of both sides, we have

㏒y = ㏒(3/2)ˣ

㏒y = x㏒(3/2)

x = ㏒y/㏒(3/2)

x = ㏒y/㏒1.5

Since we do not have logarithm of a negative number, we use y = 1.618.

So, x = ㏒y/㏒1.5

x = ㏒1.618/㏒1.5

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x = 1.19

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Learn more about exponential equation here:

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sertanlavr [38]
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