The price of a train ticket consists of an initial fee plus a constant fee per stop.

Mathematics

2 answers:

murzikaleks [220]2 years ago

8 0

Answer:

Initial Fee is $2.

Step-by-step explanation:

Given:

Stops Price (dollars)

3 6.50

7 12.50

11 18.50

Also Given:

The price of a train ticket consists of an initial fee plus a constant fee per stop.

So let the Cost of initial fee be 'x'.

Also Let the Cost of Constant fee be 'y'.

Now Equation can framed as;

$Price (P) = x + (y\times \textrm{Number of Stops})$

Now According to table;

Number of stops = 3

Price = 6.50

So equation can be framed as;

$x+3y =6.50 \ \ \ \ equation \ 1$

Also According to table;

Number of stops = 7

Price = 12.50

So equation can be framed as;

$x+7y =12.50 \ \ \ \ equation \ 2$

Now Subtracting equation 1 from equation 2 we get;

$(x+7y)-(x+3y) =12.50-6.50\\\\x+7y-x-3y=6\\\\4y =6\\\\y= \frac{6}{4}=\$1.5$

Substituting the value of y in equation 1 we get;

$x+3\times1.5=6.50\\\\x+4.5=6.50\\\\x =6.50-4.5 = \$2$

Hence Initial Fee is $2.

krek1111 [17]2 years ago

7 0

Answer:

$2.00

Step-by-step explanation:

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Which hyperbola has one focus point in common with the hyperbola (y+11)^2/(15^2)-(x-7)^2/(8^2)=1

irina [24]

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)
\end{cases}\\\\
-------------------------------\\\\
\cfrac{(y+11)^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\implies \cfrac{(y-(-11))^2}{15^2}-\cfrac{(x-7)^2}{8^2}=1\\\\
-------------------------------\\\\
c=\textit{distance from the center to either foci}\\\\
c=\sqrt{a^2+b^2}\implies c=\sqrt{15^2+8^2}\implies \boxed{c=17}$

now, if you notice, the positive fraction, is the one with the "y" variable, and that simply means, the hyperbola traverse axis is over the y-axis, so it more or less looks like the picture below.

now, from the hyperbola form, we can see the center is at (7, -11), and that the "c" distance is 17.

so, from -11 over the y-axis, we move Up and Down 17 units to get the foci, which will put them at (7, -11-17) or (7, -28) and (7 , -11+17) or (7, 6)