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3 years ago

Can someone help me with my algebra if u reply yes I’ll ask my questions

Mathematics

2 answers:

const2013 [10]3 years ago

6 0

Ok I guess?
I think I will get them right

pishuonlain [190]3 years ago

5 0

The answer is 3 x^ 2 − x + 2

hope it helps

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f of x equals 4 x over quantity x squared minus 16. Show all work to identify the asymptotes and zero of the function.

yuradex [85]

Given:

The function is

$f(x)=\dfrac{4x}{x^2-16}$

To find:

The asymptotes and zero of the function.

Solution:

We have,

$f(x)=\dfrac{4x}{x^2-16}$

For zeroes, f(x)=0.

$\dfrac{4x}{x^2-16}=0$

$4x=0$

$x=0$

Therefore, zero of the function is 0.

For vertical asymptote equate the denominator of the function equal to 0.

$x^2-16=0$

$x^2=16$

Taking square root on both sides, we get

$x=\pm \sqrt{16}$

$x=\pm 4$

So, vertical asymptotes are x=-4 and x=4.

Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.

8 0

3 years ago

Plz help me with this

AnnZ [28]

Answer: $\bold{A)\quad y=5sin\bigg(\dfrac{6}{5}x-\pi\bigg)-4}$

Step-by-step explanation:

The general form of a sin/cos function is: y = A sin/cos (Bx-C) + D where the period (P) = 2π ÷ B

In the given function, $B=\dfrac{6}{5}$ → $P=2\pi \cdot \dfrac{5}{6}=\dfrac{10\pi}{3}$

Half of that period is: $\dfrac{1}{2}\cdot \dfrac{10\pi}{3}=\large\boxed{\dfrac{5\pi}{3}}$

Calculate the period for each of the options to find a match:

$A)\quad B=\dfrac{6}{5}:\quad 2\pi \div \dfrac{6}{5}=2\pi \cdot \dfrac{5}{6}=\dfrac{5\pi}{3}\quad \leftarrow\text{THIS WORKS!}\\\\\\B)\quad B=\dfrac{6}{10}:\quad 2\pi \div \dfrac{6}{10}=2\pi \cdot \dfrac{10}{6}=\dfrac{10\pi}{3}\\\\\\C)\quad B=\dfrac{5}{6}:\quad 2\pi \div \dfrac{5}{6}=2\pi \cdot \dfrac{6}{5}=\dfrac{12\pi}{5}\\\\\\D)\quad B=\dfrac{3}{10}:\quad 2\pi \div \dfrac{3}{10}=2\pi \cdot \dfrac{10}{3}=\dfrac{20\pi}{3}$