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3 years ago

What is the equation of a hyperbola with a = 5 and c = 15? Assume that the transverse axis is horizontal.

Mathematics

1 answer:

gogolik [260]3 years ago

6 0

Answer:

$\frac{x^2}{25} -\frac{y^2}{200} =1$

Step-by-step explanation:

Recall that the equation of hyperbola centered at the origin with transverse horizontal axis has the general form:

$\frac{x^2}{a^2} -\frac{y^2}{b^2} =1\\and\,\,\,\, with\\c=\sqrt{a^2+b^2}$

since we know that a = 5 ,and c = 15,we can solve for b:

$15=\sqrt{5^2+b^2} \\15^2=25+b^2\\b^2=200\\$

then the equation of the hyperbola becomes:

$\frac{x^2}{5^2} -\frac{y^2}{200} =1\\\frac{x^2}{25} -\frac{y^2}{200} =1$

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8x^2 - 2x = 1 quadratic simplification

ziro4ka [17]

Answer:

x = -1/4 = -0.250

Step-by-step explanation:

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3 years ago

This week, we are covering relationships that can be approximated by linear equations. For instance, y = 453x + 3768 represents

lana [24]

Answer:

See explanation below.

Step-by-step explanation:

We assume that the data is given by :

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

Where X represent the cost for scholarships in thousands of dollars and y represent the cost of life for an academic semester (The data comes from the web)

We can find the least-squares line appropriate for this data.

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720$

$\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324$

$\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200$

$\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540$

$\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900$

And the slope would be:

$m=-\frac{1900}{6000}=-0.317$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60$

$\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27$

And we can find the intercept using this:

$b=\bar y -m \bar x=27-(-0.317*60)=46.02$

So the line would be given by:

$y=-0.317 x +46.02$

We have an inverse linear relationship since the slope is negative between the variables of interest.

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4 years ago

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scZoUnD [109]

Answer:

Step-by-step explanation:

The linear equation in standard form is written in the form :

ax + by = c

Therefore : 12x = -9y + 7 in standard form will be :

12x + 9y = 7

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AfilCa [17]

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