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Zolol [24]
3 years ago
8

Z3 = -1,331 Please help me​

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

z = -11

Step-by-step explanation:

ALL U HAVE TO DO IS THE OPPOSITE OF ³

WHICH IS ²

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What is the value of x?<br> O 2<br> 03<br> 06<br> 07<br> "Е
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2 years ago
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Paha777 [63]

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Step-by-step explanation:

7 0
2 years ago
45 points to answer this question
Mrac [35]

if a = b and b = c, then a = c.

In this case if y = - 7 and -7 = z then y = z  .................>(y = a, -7 = b and z = c)

Answer:  B)

transitive property of equality




7 0
3 years ago
What is the measure of Zx?
alexgriva [62]

Answer:

66

Step-by-step explanation:

180 - 105 = 75

75 + 39 = 114

180 - 114 = 66

8 0
3 years ago
Write an equation for the circle with endpoints of a diameter (9, 4) and (-3, -2)
inysia [295]
If the diameter is at (9,4) and (-3,-2), then the center of the circle is the midpoint of that segment.

and since we know that the radius of a circle is half of the diameter, whatever long that diameter segment is, the radius is half that.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }&#10;\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ 9 &,& 4~) &#10;%  (c,d)&#10;&&(~ -3 &,& -2~)&#10;\end{array}\qquad&#10;%   coordinates of midpoint &#10;\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)&#10;\\\\\\&#10;\left( \cfrac{-3+9}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{center}{(3~,~1)}

\bf -------------------------------\\\\&#10;~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;\begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;%  (a,b)&#10;&&(~ 9 &,& 4~) &#10;%  (c,d)&#10;&&(~ -3 &,& -2~)&#10;\end{array}&#10;\\\\\\&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2}&#10;\\\\\\&#10;d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}

\bf -------------------------------\\\\&#10;\textit{equation of a circle}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad &#10;radius=\stackrel{\frac{\sqrt{180}}{2}}{ r}&#10;\\\\\\&#10;(x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2}&#10;\\\\\\&#10;(x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45
7 0
3 years ago
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