Hey can someone please help explain and answer this 1 question. There's a picture. Thank you!

Mathematics

1 answer:

g100num [7]3 years ago

6 0

We'll first clear a few points.
1. A hyperbola with horizontal axis and centred on origin (i.e. foci are centred on the x-axis) has equation
x^2/a^2-y^2/b^2=1
(check: when y=0, x=+/- a, the vertices)
The corresponding hyperbola with vertical axis centred on origin has equation
y^2/a^2-x^2/b^2=1
(check: when x=0, y=+/- a, the vertices).

The co-vertex is the distance b in the above formula, such that
the distance of the foci from the origin, c satisfies c^2=a^2+b^2.
The rectangle with width a and height b has diagonals which are the asymptotes of the hyperbola.

We're given vertex = +/- 3, and covertex=+/- 5.
And since vertices are situated at (3,0), and (-3,0), they are along the x-axis.
So the equation must start with
x^2/3^2.

It will be good practice for you to sketch all four hyperbolas given in the choices to fully understand the basics of a hyperbola.

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two trains leave the station at the same time one heading west and the other East the westbound train travels at 55 miles per ho

Diano4ka-milaya [45]

Recall your d = rt, distance = rate * time

so... one train goes west and the goes the opposite way... alrite... so... notice, by the time 338 miles have been covered by both, it will have been "t" hours, and whatever "t" is, is the same amount of time the westbound train has been running as well as the eastbound train has been running.

now, let's say, since by "t" hours they've covered 338 altogether, so, if the westbound train has covered say "d" miles, then the eastbound train would have covered the slack from 338 and d, that is, "338 - d".

$\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Westbound&d&55&t\\
Eastbound&338-d&75&t
\end{array}
\\\\\\
\begin{cases}
\boxed{d}=55t\\
338-d=75t\\
----------\\
338-\boxed{55t}=75t
\end{cases}
\\\\\\
338=75t+55t\implies 338=130t\implies \cfrac{338}{130}=t\implies \stackrel{hours}{2.6}=t$

so, 2hours and 36 minutes.

8 0

3 years ago

Add 4x²y,-2xy²,-5xy²,3x²y plz explain

Vinil7 [7]

4x^2 - 2xy^2
5xy^2 +
3x^2y
_____________
12x^5y^4-2xy^2

This is so because 4+5+3 is 12, then using laws of indices to add your x and y you get x^5 and y^4

To simplify your answer to the lowest you have it in the form of
3x^2y^2(4x^2y - 2xy^2)

If you multiply this as well you get the same answer I got with the addition

3 0

3 years ago

Can someone help me find the relative maximum and minimum? please i really need help

bearhunter [10]

9514 1404 393

Answer:

relative maximum: -4
relative (and absolute) minimum: -5

Step-by-step explanation:

The curve has a relative maximum where values on either side are lower. This looks like a peak in the curve. There is one of those on the y-axis at y = -4.

The relative maximum is -4.

A relative minimum is a low point, where the curve is higher on either side. There are two of these, located symmetrically about the y-axis. The minimum appears to be about y = -5. (They might be at x = ± 1, but it is hard to tell.)

The relative minima are -5.

A minimum or maximum is absolute if no part of the curve is lower or higher. Here, the minima are absolute, while the maximum is only relative. (The left and right branches of the curve go higher than y=-4.)

_____

Identifying the points on the curve should be the easy part. Deciding what the coordinates are can be harder when the graph is like this one.

4 0

3 years ago

Which is an equation, in point-slope form, of a line that passes through the points (2,9) and (4,5)?

Elodia [21]

Only two of them are the correct answers to this question.