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MA_775_DIABLO [31]
3 years ago
12

-12m = 16 . plz its really important i need some help

Mathematics
2 answers:
Mandarinka [93]3 years ago
6 0

Answer:

-3/4

Step-by-step explanation:

-12m=16

m=12/-16

simplify

m=-3/4

babymother [125]3 years ago
3 0

Answer:

-1 and 1/3

Step-by-step explanation:

-12 x -1 1/3= 16

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What is the effect on the graph of the function f(x) = x^2 when f(x) is changed to f(x − 6)
olya-2409 [2.1K]

Answer:

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

Step-by-step explanation:

When we Add or subtract a positive constant, let say c, to input x, it would be a horizontal shift.  

For example:

Type of change               Effect on y = f(x)

y = f(x - c)                      horizontal shift: c units to right

So

Considering the function

f(x) = x^2

The graph is shown below. The first figure is representing f(x) = x^2.

Now, considering the function

\:f\left(x\right)=\left(x-6\right)^2

According to the rule, as we have discussed above, as a positive constant 6 is added to the input, so there is a horizontal shift, 6 units to the right.

The graph of \:f\left(x-6\right)=\left(x-6\right)^2 is shown below in second figure. It is clear that the graph of  \:f\left(x-6\right)=\left(x-6\right)^2  is shifted 6 units to the right as compare to the function f(x) = x^2.

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

5 0
3 years ago
Intersection point of Y=logx and y=1/2log(x+1)
GalinKa [24]

Answer:

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

The Problem:

What is the intersection point of y=\log(x) and y=\frac{1}{2}\log(x+1)?

Step-by-step explanation:

To find the intersection of y=\log(x) and y=\frac{1}{2}\log(x+1), we will need to find when they have a common point; when their x and y are the same.

Let's start with setting the y's equal to find those x's for which the y's are the same.

\log(x)=\frac{1}{2}\log(x+1)

By power rule:

\log(x)=\log((x+1)^\frac{1}{2})

Since \log(u)=\log(v) implies u=v:

x=(x+1)^\frac{1}{2}

Squaring both sides to get rid of the fraction exponent:

x^2=x+1

This is a quadratic equation.

Subtract (x+1) on both sides:

x^2-(x+1)=0

x^2-x-1=0

Comparing this to ax^2+bx+c=0 we see the following:

a=1

b=-1

c=-1

Let's plug them into the quadratic formula:

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}

x=\frac{1 \pm \sqrt{1+4}}{2}

x=\frac{1 \pm \sqrt{5}}{2}

So we have the solutions to the quadratic equation are:

x=\frac{1+\sqrt{5}}{2} or x=\frac{1-\sqrt{5}}{2}.

The second solution definitely gives at least one of the logarithm equation problems.

Example: \log(x) has problems when x \le 0 and so the second solution is a problem.

So the x where the equations intersect is at x=\frac{1+\sqrt{5}}{2}.

Let's find the y-coordinate.

You may use either equation.

I choose y=\log(x).

y=\log(\frac{1+\sqrt{5}}{2})

The intersection is (\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2}).

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