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4 years ago

NEED HELP ASAP! What is the value of x? x =

Mathematics

1 answer:

rjkz [21]4 years ago

7 0

The Intersecting Secant Theorem says

EC × ED = EB × EA

(x+4)(x+4+1) = (x+1)(x+1+11)

(x+4)(x+5)=(x+1)(x+12)

x^2 + 9x + 20 = x^2 + 13 x + 12

8 = 4x

x = 2

Check:

6(7)=42

3(14)=42, good

Answer: 2

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Compare negative two and three tenths repeating and negative eight thirds using symbols <, >, or =.

Novay_Z [31]

Answer:

$-2 < \frac{1}{3} > -\frac{8}{3}$

Note:

Three-tenths reapeating is $\frac{1}{3}$

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2 years ago

Is the line for the equation y = 1 horizontal or vertical? What is the slope of this line? Select the best answer.

Lilit [14]

Any line y=m is horizontal. The y-coordinate of any point lying on such line is m, only x-coordinates change.

If y=1, then two points on the line are (x₁,1) and (x₂,1).

$\hbox{the slope: } m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-1}{x_2-x_1}=\frac{0}{x_2-x_1}=0$

So, the slope of this and any horizontal line is 0.

The answer is a.

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3 years ago

Read 2 more answers

A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (a

Masja [62]

1. $f$ has a horizontal asymptote at $y=-4$

This means that

$\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0$

(for at least one of these limits)

2. $f$ has a vertical asymptote at $x=3$

This means that $f$ has a non-removable discontinuity at $x=3$ . Since $f$ is some rational function, there must be a factor of $x-3$ in its denominator.

3. $f$ has an $x$ -intercept at (1, 0)

This means $f(1)=0$ .

(a) With

$f(x)=\dfrac{ax+b}{x+c}$

the second point above suggests $c=-3$ . The first point tells us that

$\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0$

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if $a=-4$ so that the linear terms vanish.