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mezya [45]
3 years ago
13

Write a linear equation in

Mathematics
1 answer:
irina [24]3 years ago
7 0

\bf (\stackrel{x_1}{-11}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-11)}}}\implies \cfrac{2+5}{1+11}\implies \cfrac{7}{12}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-5)}=\stackrel{m}{\cfrac{7}{12}}[x-\stackrel{x_1}{(-11)}]\implies y+5=\cfrac{7}{12}(x+11) \\\\\\ y+5=\cfrac{7}{12}x+\cfrac{77}{12}\implies y=\cfrac{7}{12}x+\cfrac{77}{12}-5\implies y = \cfrac{7}{12}x +\cfrac{17}{12}

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Let’s take a random number such as 10, for example as the divisor of the equation.
Let’s take the dividend, or numerator of the fraction be x.
The solution of the fraction would be -2.

x/10 = -2
Or, x= -2 x 10
Or, x= -20

Therefore, -20/10 = -2
Ans: The division equation would be -20/10 = -2.
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4 years ago
Which numbers are necessary to solve this problem?
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Hello there.

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3 years ago
HELP ME AND PLS AND THXS <br> (-1, -16)<br><br><br> (3, 0)<br><br><br> (-5, 0)<br><br><br> (0, -15)
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Step-by-step explanation:

8 0
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I need help y’all s.o.s
Firdavs [7]

Please, do not present a long list of questions like this without either explanation of what kind of help you need or what you have already done. One or two problems per post is the informal limit.


I will do just the first problem for you, to get you started, and ask you to share your efforts on the next one.



x-3y≥6

-5x-3y≥-6


We are to find values of both x and y such that points (x,y) satisfy both constraints


x-3y≥6

-5x-3y≥-6.


For simplicity's sake, replace the " ≥ " sign with " = " as follows:


x-3y = 6

-5x-3y= -6


Let's solve this system thru elimination. Mult. the 2nd eqn by -1 to change the sign of all its terms:



5x + 3y = 6


and then add this result to the 1st equation:


5x + 3y = 6

x-3y = 6

-------------------

6x = 12, so x = 6. Now subst. 6 for x in x-3y = 6: 6-3y = 6. Thus, y=0.


The point at which the 2 lines cross is (6,0).


Graph both lines, using solid (not broken) lines, due to the " = " in " ≥ .


Because of the " ≥ " signs in each of the original inequalities, shade the graph ABOVE each of these solid lines. Note that you will have TWO shaded areas, which partially overlap. The solution set here is represented by the area that has been shaded twice.


To help you with the graphing:


x-3y = 6 can be rewritten as 3y = -x + 6, or y = (-1/3)x + 2. This line has a y-intercept of 2 and a slope of -1/3.


-5x-3y≥-6 can be rewritten as - 3y = 5x - 6, or y = (-5/3)x + 2. This line has a y-intercept of 2 (same as the other line) and a slope of -1/3 (different from the other one).


If you do this graphing correctly, you will find that the 2 lines intersect at (6,0).


The most efficient way to show the "solution" of this system of inequalities is to graph the two lines and shade the areas above them. Draw lines around the area that has been shaded twice.


4 0
3 years ago
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