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allochka39001 [22]
4 years ago
6

Graph the hyperbola with equation quantity x plus 3 squared divided by 25 minus the quantity of y plus 5 squared divided by 4 =

1

Mathematics
2 answers:
aleksandr82 [10.1K]4 years ago
7 0

Hello,

Please, see the graph in the attached file.

Thanks.

Sliva [168]4 years ago
7 0

Answer:

Step-by-step explanation:

As we know the general formula of the hyperbola is:

(x-h)²/a² - (y-k)²/b² = 1, where

  • (h,k) is the center
  • a is the semi-major axis
  • b is the semi-minor axis

In this situation, the equation is:

(x+3)²/25 - (y+5)²/4 = 1

=> the center is: (-3, -5)

a = 5

b = 2

  • As we know their slopes are +/- \frac{b}{a} so

-5 = +2/5 (-3) + b ---> b= -3.8

-5 = -2/5 (-3) + b ---> b= -6.2

Hence, you plot the equations y=2/5x -3.8  and y = -2/5 x - 6.2 by assigning values of x and plotting them against y as the attached photo.

Hope it will find you well.

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3 0
3 years ago
Slope=-3 goes through (1,2) whats the equation
dedylja [7]

Hi student, let me help you out! :)

..............................................................................................................................................

We are asked to find the equation of the line, using the given information:

\star~\mathrm{-3}, which is the slope

\star~\mathrm{(1,2)} which is a point that the line contains

\triangle~\fbox{\bf{KEY:}}

  • The slope is the number before x, in the form \mathtt{y=mx+b}.

Since "m" is the slope, and we're given what the slope is, we can substitute -3 for m:

\dag~\mathtt{y=-3x+b}

|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Now, we need to find the y intercept, "b", of the line.

Here's how it's done:

Remember the point that the line contains, (1,2)?

Well, its y-coordinate is 2; so what we do is substitute 2 for y:

\dag~\mathtt{2=-3x+b}

Also, its x-coordinate is 1, so we substitute 1 for x:

\dag~\mathtt{2=-3(1)+b}

Now, solve for b:

\dag~\mathtt{2=-3+b}

Add 3 to both sides:

\dag~\mathtt{2+3=b}

Thus, we have

\dag~\mathtt{5=b}

Thus, the equation of the line is:

\bigstar\underline{\boxed{\mathtt{y=-3x+5}}}

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students :)

\overline{\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~}}

3 0
2 years ago
convert 13542 base six to base ten. I'm in math20 at Folsom Lake College and i'm trying to figure out this question. I had most
VikaD [51]
Converting to base 10 can be done easily by writing a sort of digit expansion. In base 6, this number literally means

13542_6=1\times6^4+3\times6^3+5\times6^2+4\times6^1+2\times6^0

Simplifying this gives the value in base 10.

13542_6=1296+3\times216+5\times36+4\times6+2
13542_6=2150_{10}
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4 years ago
How would you determine the axis of symmetry in order to graph x^2 = 8y^2.
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The picture illustrates the definition. The point P is a typical point on the parabola so that its distance from the directrix, PQ, is equal to its distance from F, PF. The point marked V is special. It is on the perpendicular line from F to the directix. This line is called the axis of symmetry of the parabola or simply the axis of the parabola and the point V is called the vertex of the parabola. The vertex is the point on the parabola closest to the directrix.

Finding the equation of a parabola is quite difficult but under certain cicumstances we may easily find an equation. Let's place the focus and vertex along the y axis with the vertex at the origin. Suppose the focus is at (0,p). Then the directrix, being perpendicular to the axis, is a horizontal line and it must be p units away from V. The directrix then is the line y=-p. Consider a point P with coordinates (x,y) on the parabola and let Q be the point on the directrix such that the line through PQ is perpendicular to the directrix. The distance PF is equal to the distance PQ. Rather than use the distance formula (which involves square roots) we use the square of the distance formula since it is also true that PF2 = PQ2. We get

<span>(x-0)2+(y-p)2 = (y+p)2+(x-x)2. 
x2+(y-p)2 = (y+p)2.</span>If we expand all the terms and simplify, we obtain<span>x2 = 4py.</span>

Although we implied that p was positive in deriving the formula, things work exactly the same if p were negative. That is if the focus lies on the negative y axis and the directrix lies above the x axis the equation of the parabola is

<span>x2 = 4py.</span><span>The graph of the parabola would be the reflection, across the </span>x<span> axis of the parabola in the picture above. A way to describe this is if p > 0, the parabola "opens up" and if p < 0 the parabola "opens down".</span>

Another situation in which it is easy to find the equation of a parabola is when we place the focus on the x axis, the vertex at the origin and the directrix a vertical line parallel to the y axis. In this case, the equation of the parabola comes out to be

<span>y2 = 4px</span><span>where the directrix is the verical line </span>x=-p and the focus is at (p,0). If p > 0, the parabola "opens to the right" and if p < 0 the parabola "opens to the left". The equations we have just established are known as the standard equations of a parabola. A standard equation always implies the vertex is at the origin and the focus is on one of the axes. We refer to such a parabola as a parabola in standard position.

Parabolas in standard positionIn this demonstration we show how changing the value of p changes the shape of the parabola. We also show the focus and the directrix. Initially, we have put the focus on the y axis. You can select on which axis the focus should lie. Also you may select positive or negative values of p. Initially, the values of p are positive. 

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A snail crawled of a centimeter in 2 minutes. At that rate, how far could the snail crawl in 1
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Answer:30 centimeters in 1 minute

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