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Marrrta [24]
3 years ago
10

24y - 22 = 4 ( 6y - 6 )

Mathematics
1 answer:
Pie3 years ago
5 0
I don’t think their is a solution to this equation
because if you expand the second half it is= 24y-24 which would make the equation
- 24y-22=24y-24
and because the number next to the y is the same on both sides, no matter what y is if we subtract different numbers from each side we will never get the same value for each side of the =
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Finish the value of d 2d - 5 = 17
Mkey [24]
D is 11

2d - 5 = 17                          add 5 to both sides

2d = 22                               divide by 2

d = 11                                 answer
8 0
3 years ago
hi:) I’m so sorry , i uploaded this question earlier but I think the picture couldn’t load. Yea I need help with this question,
love history [14]

Answer:

(ii) \frac{1}{2}  -  \frac{1}{3}

(iii) \frac{25}{51}

Step-by-step explanation:

Please see the attached picture for full solution.

7 0
3 years ago
Find the limit. (let g and h represent arbitrary real numbers. if an answer does not exist, enter dne.) lim x→∞ x2 + gx − x2 + h
seraphim [82]
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8 0
4 years ago
Find the equation of the circle that has a diameter with endpoints located at (7, 3) and (7, –5).
posledela
So hmm check the picture below, that's about the circle and the endpoints, but notice, the endpoints make up a segment, namely the diameter of the circle, well.... let's see how long that is, because, the radius is half the diameter

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad 
%  (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{(7-7)^2+(-5-3)^2}\implies d=\sqrt{0+(-8)^2}\implies d=8
\\\\\\
\textit{the radius is half that, so is }\boxed{r=4}

now.. hmmm notice, the midpoint of the diameter, is the center of the circle, let's check that one out

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ 7}}\quad ,&{{ 3}})\quad 
%  (c,d)
&({{ 7}}\quad ,&{{ -5}})
\end{array}\qquad 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 7+7}}}{2}\quad ,\quad \cfrac{{{ -5}} + {{ 3}}}{2} \right)\implies \left( \cfrac{14}{2}\ ,\ \cfrac{-2}{2} \right)\implies \boxed{(7,-1)}

now.. that we know what the center is, and what the radius is, well

\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k\quad }})\qquad 
radius=&{{ r}}\\
&7&-1&4
\end{array}

5 0
3 years ago
Crossing the River
viktelen [127]

Answer:

The answer you are looking for is 5

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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