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3 years ago

Which statement is true about the graphed function?

Mathematics

2 answers:

Semmy [17]3 years ago

6 0

A function is positive if it lies above the x axis.

This function is positive before x = -4, i.e. over the interval $(-\infty, -4)$

It is negative elsewhere, i.e. over the interval $(-4, \infty)$

So, the only correct option is the second one.

grigory [225]3 years ago

3 0

Answer:

D.F(x) >0 over the interval ( $-\infty,-4)$

Step-by-step explanation:

We have to find the true statement about graphed function.

A.F(x) <0 over the interval ( $-\infty,4$ )

It is not true because in given graph the value of F(x) < 0 in interval ( $-4,\infty,)$ and F(x)> 0 in interval ( $-\infty,-4)$ .

B.F(x)<0 over the interval ( $-\infty,-3)$

It is not true because F(x)> 0 in interval ( $-\infty,-4)$ .

C.F(x) >0 over the interval $(-\infty,-3)$

It is not true because F(x) < 0 when -4<x<-3

D.F(x) >0 over the interval ( $-\infty,-4)$

It is true because F(x) >0 in interval $(-\infty,-4)$

Hence, option D is true.

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Solve: (x ÷ 4) + 8 = 38

Rainbow [258]

First you have to rewrite the equation:
(x ÷ 4) + 8 = 38

Second you have to write the division as fraction:
(¼x) + 8 = 38

Third you have to take off the unecessary bracket:

¼x + 8 = 38

Fourth multiply 8 and 38 by 4
x + 32 = 152

Fifth step move the constant to the right hand and change the sign.
x = 152 - 32

Then subtract:

x = 120

Hope this helps :))

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3 years ago

Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x > 0)

vfiekz [6]

Answer:

Step-by-step explanation:

given a point $(x_0,y_0)$ the equation of a line with slope m that passes through the given point is

$y-y_0 = m(x-x_0)$ or equivalently

$y = mx+(y_0-mx_0)$ .

Recall that a line of the form $y=mx+b$ , the y intercept is b and the x intercept is $\frac{-b}{m}$ .

So, in our case, the y intercept is $(y_0-mx_0)$ and the x intercept is $\frac{mx_0-y_0}{m}$ .

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph $(x_0,\frac{1}{x_0})$ . Which means that $y_0=\frac{1}{x_0}$