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MrMuchimi
2 years ago
10

The functions f(x) and g(x) are shown on the graph.

Mathematics
2 answers:
katrin [286]2 years ago
7 0

Answer:

g(x)=\log_3(x+4)

Step-by-step explanation:

<u>Translations</u>

For a > 0

f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}

f(x-a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units right}

f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}

f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}

<u>Parent function:</u>

  f(x)=\log_3x

From inspection of the graph:

  • The x-intercept of f(x) is (1, 0)
  • The x-intercept of g(x) is (-3, 0)

If there was a vertical translation, the end behaviors of both g(x) and f(x) would be the same in that the both functions would be increasing from -∞ in quadrant IV.

As the x-intercepts of both functions is different, and g(x) increases from -∞ in quadrant III, this indicates that there has been a <u>horizontal translation</u> of <u>4 units to the left</u>.

Therefore:

g(x)=f(x+4)=\log_3(x+4)

<u>Further Proof</u>

Logs of zero or negative numbers are <u>undefined</u>.

From inspection of the graph, x=-3 is part of the domain of g(x).

Therefore, input this value of x into the answer options:

 g(-3)=\log_3(-3)-4\implies undefined

 g(-3)=\log_3(-3)+4\implies undefined

 g(-3)=\log_3(-3-4)=\log_3(-7)=\implies undefined

 g(-3)=\log_3(-3+4)=\log_3(1)=0

Hence proving that g(x)=\log_3(x+4)

Dafna11 [192]2 years ago
7 0

g(x) is the translated edition of f(x)

  • f(x)=log_3x

Let y=f(x)

  • y=log_3x

Now

The graph is shifted 4 units left means change in x axis.

  • y=log_3(x+4)

No change in y

So

  • g(x)=log_3(x+4)

Option D

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Solve the system of equations.<br><br><br><br> −2x+5y =−35<br> 7x+2y =25
Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}

This system of equations can be solved in three different ways:

  1. Graphing the equations (method used)
  2. Substituting values into the equations
  3. Eliminating variables from the equations

<u>Graphing the Equations</u>

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is \text{y = mx + b}.

Equation 1 is -2x+5y = -35. We need to isolate y.

\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7

Equation 1 is now y=\frac{2}{5}x-7.

Equation 2 also needs y to be isolated.

\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}

Equation 2 is now y=-\frac{7}{2}x+\frac{25}{2}.

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}

\bullet \ \text{For x = 0,}

\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7

\bullet \ \text{For x = 1,}

\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}

\bullet \ \text{For x = 2,}

\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}

\bullet \ \text{For x = 3,}

\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}

\bullet \ \text{For x = 4,}

\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}

\bullet \ \text{For x = 5,}

\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5

Now, we can place these values in our table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

As we can see in our table, the rate of decrease is -\frac{2}{5}. In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract -\frac{2}{5} from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be y=-\frac{7}{2}x+\frac{25}{2}. Therefore, we just use the same process as before to solve for the values.

\bullet \ \text{For x = 0,}

\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}

\bullet \ \text{For x = 1,}

\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9

\bullet \ \text{For x = 2,}

\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}

\bullet \ \text{For x = 3,}

\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2

\bullet \ \text{For x = 4,}

\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}

\bullet \ \text{For x = 5,}

\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5

And now, we place these values into the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1                  Equation 2

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}                 \begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

Therefore, using this data, we have one solution at (5, -5).

4 0
3 years ago
What is this answer 9x2000​
r-ruslan [8.4K]
18,000


just use a calculator bro
6 0
3 years ago
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Find the value of C.
Elan Coil [88]

Answer:

C

Step-by-step explanation:

We already know that a right angle is 90° and where C is in the picture shows us an obtuse angle making it beleive that it must be above 90°, which leaves us with C as our final answer and is the only answer that makes sense to be correct. Hopefully this helped you.

5 0
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In the data set shown below, what is the value of the quartiles? {42, 43, 44, 44, 48, 49, 50} A. Q1 = 43.5; Q2 = 44; Q3 = 49 B.
ANTONII [103]

Answer:

C

Step-by-step explanation:

I don't actually know how to explain it but,

Q1=43.5

Q2=44

Q3=48.5

Sorry, I can't help more.

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since x + x + x is 3x.

Or you can be creative

(6x+4)/2

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