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KonstantinChe [14]
3 years ago
11

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±five divided by four times

x..
Mathematics
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

The equation of the hyperbola in standard form is

\frac{y^{2}}{100}-\frac{4x^{2}}{625}=1

Step-by-step explanation:

* We will take about the standard form equation of the hyperbola

- If the given coordinates of the vertices (0 , a) and (0 , -a)

∴ The transverse axis is the y-axis. (because x = 0)

- If the given asymptotes at y = ± (b/a) x

∴  Use the standard form  ⇒ y²/a² - x²/b² = 1

* Lets use this to solve our problem

∵ The vertices are (0 , 10) and (0 , -10)

∴ a = ±10

∴ a² = 100

∵ The asymptotes at y = ± 5/4 x

∴ ± 5/4 = ± b/a

∵ a = ± 10

∴ ± 5/4 = ± b/10 ⇒ using cross multiplication

∴± (4b) = ± (5 × 10) = ± 50 ⇒ divide both sides by 4

∴ b = ± 25/2

∴ b² = 625/4

* Now Lets write the equation

* y²/100 - x²/(625/4) = 1

∵ x² ÷ 625/4 = x² × 4/625 = (4x²/625)

∴ y²/100 - 4x²/625 = 1

* The equation of the hyperbola in standard form is

  \frac{y^{2}}{100}-\frac{4x^{2}}{625}=1

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<span>√p − 1 − √p = 1
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What is the area of this figure? <br>​
polet [3.4K]
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2 years ago
Integrate 1 - x / x(x2 + 1) d x by partial fractions.
solniwko [45]

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log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x

Step-by-step explanation:

step 1:-   by using partial fractions

[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }......(1)

<u>step 2:-</u>

solving on both sides

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substitute x =0 value in equation (2)

1=A(1)+0

<u>A=1</u>

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0 = A+B

0 = 1+B

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comparing x co-efficient on both sides (in equation 2)

<u>-</u>1  =  C

<u>step 3:-</u>

substitute A,B,C values in equation (1)

now  

\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 }  d x -\int\limits \frac{1}{x^{2}+1 }  d x

by using integration formulas

i)  by using \int\limits \frac{1}{x}   d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\.....(b)

\int\limits tan^{-1}x  dx =\frac{1}{1+x^{2} } +C.....(c)

<u>step 4:-</u>

by using above integration formulas (a,b,and c)

we get answer is

log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x

6 0
3 years ago
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ivanzaharov [21]

Answer:

Infinite pairs of numbers

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8 and -8

Step-by-step explanation:

Let x³ and y³ be any two real numbers. If the sum of their cube roots is zero, then the following must be true:

\sqrt[3]{x^3}+ \sqrt[3]{y^3}=0\\ \sqrt[3]{x^3}=- \sqrt[3]{y^3}\\x=-y

Therefore, any pair of numbers with same absolute value but different signs fit the description, which means that there are infinite pairs of possible numbers.

Examples: 1 and -1; 8 and -8; 27 and -27.

8 0
3 years ago
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