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Ostrovityanka [42]
2 years ago
12

5) The graph of the parent linear function f(x) = x is shown in black on the coordinate grid. Write the function that represents

this function with the indicated parameter changes.
Slope (m) increased, y-intercept (b) unchanged.
Slope (m) decreased, y-intercept (b) unchanged.
Slope (m) unchanged, y-intercept (b) increased.
Slope (m) unchanged, y-intercept (b) decreased.
Mathematics
1 answer:
djyliett [7]2 years ago
8 0

Answer:

(m) increased, (b) unchanged.     g(x)

(m) decreased, (b) unchanged.   m(x)

(m) unchanged, (b) increased.      h(x)

(m) unchanged, (b) decreased.    n(x)

f(x): m = -1/2  b = 2

g(x): m = 1/3  b = -3

h(x): m = 2  b = 0

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Graph the image of this figure after a dilation with a scale factor of 12 centered at the point (2, 2) . Use the polygon tool to
Alla [95]

Answer:

Labelled the given figure triangle as A, B and C.

In triangle ABC;

The coordinates of ABC are:

A = (0, 4)

B = (-2, 10) and

C = (-6, 8)

To find the image of this figure after a dilation with a scale factor of \frac{1}{2} centered at the point O=(2, 2).

The rule of dilation with scale factor \frac{1}{2} centered at the point (2, 2) is given by;

(x, y) \rightarrow (\frac{1}{2}(x-2) +2 , \frac{1}{2}(y-2) +2)

or

(x, y) \rightarrow (\frac{1}{2}x+1 , \frac{1}{2}y+1)

The coordinate of the image of the figure after dilation are;

A(0, 4) \rightarrow A'(\frac{1}{2}(0)+1 , \frac{1}{2}(4)+1) = A'(1, 3)

B(-2, 10) \rightarrow B'(\frac{1}{2}(-2)+1 , \frac{1}{2}(10)+1) = B'(0, 6)

C(-6, 8) \rightarrow C'(\frac{1}{2}(-6)+1 , \frac{1}{2}(8)+1) = C'(-2, 5)

You can see the graph as shown below in the attachment:





5 0
3 years ago
Find the area. Simplify your answer. 2x-2 3x​
suter [353]

Answer:

i think 5x^2-5

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(3X)(2X-2)

5X^2-5

5 0
3 years ago
Read 2 more answers
Number 1d please help me analytical geometry
lesantik [10]
For a) is just the distance formula

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ x}}\quad ,&{{ 1}})\quad 
%  (c,d)
B&({{ -4}}\quad ,&{{ 1}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\sqrt{8} = \sqrt{({{ -4}}-{{ x}})^2 + (1-1)^2}
\end{array}
-----------------------------------------------------------------------------------------
for b)  is also the distance formula, just different coordinates and distance

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -7}}\quad ,&{{ y}})\quad 
%  (c,d)
B&({{ -3}}\quad ,&{{ 4}})
\end{array}\ \ 
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
4\sqrt{2} = \sqrt{(-3-(-7))^2+(4-y)^2}
\end{array}
--------------------------------------------------------------------------
for c)  well... we know AB = BC.... we do have the coordinates for A and B
so... find the distance for AB, that is \bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
A&({{ -3}}\quad ,&{{ 0}})\quad 
%  (c,d)
B&({{ 5}}\quad ,&{{ -2}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=\boxed{?}

\end{array}

now.. whatever that is, is  = BC, so  the distance for BC is

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
B&({{ 5}}\quad ,&{{ -2}})\quad 
%  (c,d)
C&({{ -13}}\quad ,&{{ y}})
\end{array}\qquad 
%  distance value
\begin{array}{llll}

d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\
d=BC\\\\
BC=\boxed{?}

\end{array}

so... whatever distance you get for AB, set it equals to BC, BC will be in "y-terms" since the C point has a variable in its ordered points

so.. .solve AB = BC for "y"
------------------------------------------------------------------------------------

now d)   we know M and N are equidistant to P, that simply means that P is the midpoint of the segment MN

so use the midpoint formula

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
M&({{-2}}\quad ,&{{ 1}})\quad 
%  (c,d)
N&({{ x}}\quad ,&{{ 1}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=P
\\\\\\


\bf \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(1,4)\implies 
\begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=1\leftarrow \textit{solve for "x"}\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2}=4
\end{cases}

now, for d), you can also just use the distance formula, find the distance for MP, then since MP = PN, find the distance for PN in x-terms and then set it to equal to MP and solve for "x"


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