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2 years ago

An elevator went up 15 floors, down 9 floors, up 11 floors and down 19 floors. Find the net change

Mathematics

2 answers:

mariarad [96]2 years ago

8 0

Answer:

The net change would be -2.

Step-by-step explanation:

The upward change of elevator is represented by positive sign while the downward change represented by negative sign.

Given,

First the elevator goes up 15 floors,

change = + 15,

Then it goes 9 floors down,

change = -9,

Then it goes 11 floors up,

Change = + 11,

Finally, it goes 19 floor down,

Change = - 19

So, the net change = sum of total changes = 15 - 9 + 11 - 19 = -2

Slav-nsk [51]2 years ago

3 0

15u + 11u - 9d - 19d = -2d
It went up 26, down 28, a net change of down 2.

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Tickets to a show cost $5.50 for adults and $4.25 for students. A family is purchasing 2 adult tickets and 3 student tickets. If

larisa [96]

Answer:

Step-by-step explanation:

The cost of an adult ticket to the show is $5.5

The cost of a student ticket to the show is $4.25

A family is purchasing 2 adult tickets and 3 student tickets. This means that the total cost of 2 adult tickets would be

2 × 5.5 = $11

It also means that the total cost of 3 student tickets would be

3 × 4.25 = $12.75

The total cost of 2 adult tickets and 3 student tickets would be

11 + 12.75 = $23.75

If the family pays $25, the exact amount of change they should receive is

25 - 23.75 = $1.25

5 0

2 years ago

silvia compro la misma licuadora que daniela, que costaba $355.5, pero como abonó con tarjeta de credito, le recargaron un 16%,

Debora [2.8K]

Answer:

Costo final= $412.38

Step-by-step explanation:

Dada la siguiente información:

Costo inicial= $355.5

Recargo de la tarjeta= 16% = 0.16

<u>Para calcular el costo final que debe pagar Silvia, debemos usar la siguiente información:</u>

Costo final= costo inicial*(1 + recargo)

Costo final= 355.5*1.16

Costo final= $412.38

4 0

1 year ago

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)