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3 years ago

Solve for x: 4(x + 2) = 3(x − 2)

Mathematics

2 answers:

Oliga [24]3 years ago

5 0

Answer:

x = -14

Step-by-step explanation:

4x + 8 = 3x - 6

4x = 3x - 14

x = -14

Verizon [17]3 years ago

4 0

ANSWER BELOW

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

Combine like terms

4x + 8 = 3x -6

Subtract 3x from both sides

X + 8 = -6

Subtract 8 from both sides

X = -14

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Find the vertical and horizontal asymptote if they exist

tresset_1 [31]

Hello!

Vertical asymptotes are determined by setting the denominator of a rational function to zero and then by solving for x.

Horizontal asymptotes are determined by:

1. If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

2. If the degree of the numerator = degree of denominator, then y = leading coefficient of numerator / leading coefficient of denominator is the horizontal asymptote.

3. If degree of numerator > degree of denominator, then there is an oblique asymptote, but no horizontal asymptote.

To find the vertical asymptote:

2x² - 10 = 0

2(x² - 5) = 0

(x - √5)(x + √5) = 0

x = √5 and x = -√5

Graphing the equation, we realize that x = -√5 is not a vertical asymptote, so therefore, the only vertical asymptote is x = √5.

To find the horizontal asymptote:

If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

Therefore, the horizontal asymptote of this function is y = 0.

Short answer: Vertical asymptote: x = √5 and horizontal asymptote: y = 0

7 0

3 years ago

Find the rat job of rmt

Nutka1998 [239]

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5 0

3 years ago

Tan(x+pi) + 2sin(x+pi)=0<br><br>can you help prove this and make the left side equal the right?

vladimir1956 [14]

This is not an identity but an equation. You're supposed to find the values of $x$ for which this is true. It is *not* true for all values of $x$ .

To see why:

$\tan x$ has period $\pi$ , which means $\tan(x+\pi)=\tan x$ .

The angle sum identity for $\sin x$ says that

$\sin(x+\pi)=\sin x\cos \pi+\cos x\sin\pi=-\sin x$

$\tan(x+\pi)+2\sin(x+\pi)=\tan x-2\sin x$

By definition of $\tan x$ ,

$\tan x=\dfrac{\sin x}{\cos x}$

and so

$\tan x-2\sin x=\sin x\left(\dfrac1{\cos x}-2\right)$

which is only 0 if either $\sin x=0$ (which only happens for certain values of $x$ ) or $\dfrac1{\cos x}-2=0\implies\cos x=\dfrac12$ (which also only happens for certain values of $x$ ). It's these values of $x$ you want to find.

$\sin x=0$ whenever $x$ is a multiple of $\pi$ , i.e. $x=n\pi$ for any integer $n$ .

If $0\le x$ , then $\cos x=\dfrac12$ is true for $x=\dfrac\pi3$ and $x=\dfrac{5\pi}3$ . Then to account for all other possible values, we add a multiple of $2\pi$ , so that $x=\dfrac\pi3+2n\pi$ or $x=\dfrac{5\pi}3+2n\pi$ for integers $n$ .