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3 years ago

ASAPPPPPPPPP

Mathematics

2 answers:

maxonik [38]3 years ago

8 0

Answer:the answer is C. I pray for you all I hope you graduate!

Step-by-step explanation:

Neporo4naja [7]3 years ago

3 0

Answer:

The answer is D

3(x+4)(x-i)(x+i)(x-2)

Step-by-step explanation:

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Write a equation of a hyperbola given the foci and the asymptotes

professor190 [17]

Solution:

The standard equation of a hyperbola is expressed as

$\begin{gathered} \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\text{ \lparen parallel to the x-axis\rparen} \\ \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ \lparen parallel to the y-axis\rparen} \end{gathered}$

Given that the hyperbola has its foci at (0,-15) and (0, 15), this implies that the hyperbola is parallel to the y-axis.

Thus, the equation will be expressed in the form:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\text{ ----equation 1}$

The asymptote of n hyperbola is expressed as

$y=\pm\frac{a}{b}(x-h)+k$

Given that the asymptotes are

$y=\frac{3}{4}x\text{ and y=-}\frac{3}{4}x$

This implies that

$a=3,\text{ and b=4}$

To evaluate the value of h and k,

$undefined$

3 0

1 year ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7Bx%2B7%7D%20%2B%5Cfrac%7B4%7D%7Bx-8%7D" id="TexFormula1" title="\frac{3}{x+7}

sattari [20]

{7x+4}/{x^2-x-56}

using common denominators

6 0

2 years ago

Find the 4th term <br><br> Thank you in advance

shepuryov [24]

B1 = 2
b2 = (b1)^2 + 1 = 2^2 + 1 = 5
b3 = (b2)^2 + 1 = 5^2 + 1 = 26

b4 = (b3)^2 + 1 = 26^2 + 1 = 676+1=<span>677</span>

7 0

3 years ago

Read 2 more answers

Reptile [31]

Answer:

The real zeros of f(x) are x = 0.3 and x = -3.3.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem, we have that:

$f(x) = x^{2} + 3x - 1$

$a = 1, b = 3, c = -1$

$\bigtriangleup = 3^{2} - 4*1*(-1) = 13$

$x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3$

$x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3$

The real zeros of f(x) are x = 0.3 and x = -3.3.

7 0

3 years ago

What side is this a isosceles a scalene or equilateral

aksik [14]

The answer is a scalene

3 0

3 years ago