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leonid [27]
3 years ago
13

Check whether the given value is a solution to the equation.

Mathematics
1 answer:
Softa [21]3 years ago
6 0

Answer:

FALSE. The correct answer is y = - 30.

Explanation:

3*(30+8) = 2(30)-6

3*38 = 60-6

114 = 54

<em><u>The correct answer is y = -30.</u></em> (Negative 30)

3*(-30+8) = 2(-30)-6

-66 = -66

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How do you do number 16?
MA_775_DIABLO [31]
The answer to this would be 1/5. 
hope this helps
7 0
3 years ago
Need help
I am Lyosha [343]

Answer:

see in the graph, the circle has: the centre I(3; -2)

r = 3

=> the equation: (x - 3)² + (y + 2)² = 9

5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!
Brilliant_brown [7]

-2<em>x</em> + 6<em>y</em> = -38

3<em>x</em> - 4<em>y</em> = 32

To solve by elimination, multiply the top equation by 3 and the bottom equation by 2.

3(-2<em>x</em> + 6<em>y</em> = -38) --> -6<em>x</em> + 18<em>y</em> = -114

2(3<em>x</em> - 4<em>y</em> = 32) --> 6<em>x</em> - 8<em>y</em> = 64

Add the equations.

-6<em>x </em>+ 18<em>y</em> = -114

6<em>x</em> - 8<em>y</em> = 64

+_____________

0 + 10<em>y</em> = -50

10<em>y</em> = -50

<em>y</em> = -5

Substitute -5 for y into one of the original equations to find x.

3<em>x</em> - 4<em>y</em> = 32

3<em>x</em> - 4(-5) = 32

3<em>x</em> + 20 = 32

3<em>x</em> = 12

<em>x</em> = 4

Check work by plugging the <em>x</em>- and <em>y</em>-values into both of the original equations.

-2<em>x</em> + 6<em>y</em> = -38

-2(4) + 6(-5) = -38

-8 - 30 = 38

38 = 38


3<em>x</em> - 4<em>y</em> = 32

3(4) - 4(-5) = 32

12 + 20 = 32

32 = 32

Answer:

<em>x</em> = 4 and <em>y</em> = -5; (4, -5).

8 0
3 years ago
Read 2 more answers
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \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}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
4(x-4)-3x=10 what is the answer
max2010maxim [7]

Answer:

26

Step-by-step explanation:

4(x-4)-3x=10

  • Rule = a(b + c) = ab + ac
  • Rule = a(b - c) = ab - ac

4(x-4) = 4x - 16

4(x-4)-3x=10

4x - 16 - 3x = 10

x - 16 = 10

x -16 +16 = 10 +16

x = 26

Hope this helps ^-^

6 0
3 years ago
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