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Lena [83]
3 years ago
15

Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one sam

e-day ticket is
$35
. For one performance,
40
advance tickets and
30
same-day tickets were sold. The total amount paid for the tickets was
$1250
. What was the price of each kind of ticket?
Mathematics
2 answers:
icang [17]3 years ago
8 0
Advance=x

Same.day=y

x+y=35 so x=35-y

40x+30y=1250

40(35-y)+30y=1250

-10y=-150

y=15

x+15=35

x=20

Cost for Advance ticket: $20
Cost for Same-day ticket: $15



Neporo4naja [7]3 years ago
6 0
Let's set up variables and equations for things that we know from the question, with a standing for advance tickets, and s standing for same day tickets. 

40a + 30s = 1250 
a + s = 35

Let's solve it through substitution. Isolate either variable, a or s. I'll choose to isolate a. Subtract s from both sides.

a = -s + 35

Plug the new knowledge into the first equation, since we now know the value of a

40(-s +35) + 30s = 1250.

Distribute the 40 to the -s and 35

-40s + 1400 + 30s = 1250

Combine the negative and positive s

-10s + 1400 = 1250

Subtract 1400 from both sides.

-10s = -150

Divide both sides by -10

s = 15

The total price of an advanced ticket and same-day ticket is 35, and we know the value of a same-day ticket, so the price of an advance ticket must be 20. 

The final answers are:
advanced ticket = 20
same day ticket = 15


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Step-by-step explanation:

1 4| 5 0 4

2 8 0 20 <---- 20 x 14 = 280

2 2 4 <---- 504 - 280 = 224

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Step-by-step explanation:

Given the following information:

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Hence, we have:

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Answer:

\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x

Rewrite 9 as 3²  and rewrite the 3/2 exponent as square root to the power of 3:

\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x

<u>Integration by substitution</u>

<u />

<u />\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}

\textsf{Let }x=3 \sin \theta

\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}

Find the derivative of x and rewrite it so that dx is on its own:

\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta

\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta

<u>Substitute</u> everything into the original integral:

\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & =  \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}

Take out the constant:

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\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}

\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta = \dfrac{1}{9} \tan \theta+\text{C}

\textsf{Use the trigonometric identity}: \quad \tan \theta=\dfrac{\sin \theta}{\cos \theta}

\implies \dfrac{\sin \theta}{9 \cos \theta} +\text{C}

\textsf{Substitute back in } \sin \theta=\dfrac{x}{3}:

\implies \dfrac{x}{9(3 \cos \theta)} +\text{C}

\textsf{Substitute back in }3 \cos \theta=\sqrt{9-x^2}:

\implies \dfrac{x}{9\sqrt{9-x^2}} +\text{C}

Learn more about integration by substitution here:

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