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MissTica
3 years ago
6

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 :))
5 0
2 years ago
Angelica did an experiment where she measured the temperature outside her house each day for two weeks. She found that on the fi
USPshnik [31]
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3 0
3 years ago
Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x &gt; 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}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0
3 years ago
Triangle L M Q is cut by perpendicular bisector L N. Angle N L Q is 32 degrees and angle L M N is 58 degrees.
Serga [27]

Answer:

Yes

Step-by-step explanation:

ΔMNL ≅ ΔQNL  by ASA or AAS

by ASA

Proof:

∠ LNM = ∠LNQ    =90

LN = LN   {Common}

∠MLN = ∠QLN     {LN bisects ∠ L}

By AAS

∠Q + ∠QLN + ∠LNQ = 180  {Angle sum property of triangle}

∠Q + 32 + 90 = 180

∠Q  + 122 = 180

∠Q = 180 -122 =

∠Q = 58

∠Q = ∠M

∠MNL =∠QNL = 90

LN = LN {common side}

8 0
3 years ago
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aniked [119]
I think it’s going to be 7! Because since the other triangle is 1 and 3 it’s just 2 more so I just added to more to 5 getting me 7!! I think lol
3 0
2 years ago
Read 2 more answers
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