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3 years ago

7

Quadrilateral GHIJ has coordinates G(5, 1), H(4, -3), I(6, -1), and J(2, 1). If the quadrilateral is reflected over the line y =

-x, what are the coordinates of G'?

1 answer:

Grace [21]3 years ago

6 0

Answer:

Where is the diagram

Step-by-step explanation:

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Find the gcf of each pair of number 24 and 56

pshichka [43]

8is the correct answer

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3 years ago

HELP PLS GIVING BRAINLIEST

Setler79 [48]

For this case we have by definition, that the equation of a line in the slope-intersection form is given by:

$y = mx+b$

Where:

m: It's the slope

b: It is the cutoff point with the y axis

We need two points through which the line passes to find the slope:

$(0,1)\\(1, -2)$

We found the slope:

$m = \frac {y2-y1} {x2-x1}\\m = \frac {-2-1} {1-0} = \frac {-3} {1} = - 3$

So, the equation is of the form:

$y = -3x + b$

We substitute a point to find "b":

$1 = -3 (0) + b\\1 = b$

Finally, the equation is:

$y = -3x+1$

Answer:

Option C

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3 years ago

ASAP <br><br> -JUST ANSWER- PLSSSSS

Leviafan [203]

Answer:

D 100 revolution is your answer

5 0

3 years ago

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"

7 0

3 years ago

One spinner has three equally sized sectors numbered 1, 2, and 3. A second spinner has two equally sized sectors labeled Top (T)

SSSSS [86.1K]

I would say, well, if there are 3 sides that are equal on spinner 1. They are equal. But, we are focusing on spinner two that has two sides. If there is only 2 equal sides then, they would have to be straight across from each other. So that is 180. 360 is full circle. take away 180 to get answer.
360-180=180
I hope this helped. And sorry if it is not the correct answer. But that is how I would of answered it

4 0

3 years ago

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