Considering the definition of the absolute value function, the expression is written as -7m + 56.
<h3>What is the definition of the absolute value function?</h3>
The absolute value function is defined as follows:
.
That is, where the expression is negative, it is multiplied by -1, while where it is not negative the expression is kept.
In this problem, the expression is:
|7m - 56|
For m < 8, we have that one example is m = 0:
7(0) - 56 = -56.
The expression is negative, hence the signal is changed, that is:
|7m - 56| = -(7m - 56) = -7m + 56.
More can be learned about the absolute value function at brainly.com/question/24734454
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Answer:
C. Perpendicular Lines
D. Intersecting Lines
Step-by-step explanation:
Intersecting lines are a pair of lines, line segments or rays are intersecting if they have a common point.
Perpendicular lines are a point where two city roads intersect.
Hope this helps. I am sorry if you get it wrong.
Answer:
C. (3, 3)
Step-by-step explanation:
Hello! I can help you! First things first, because they are both the same angles on opposite sides, let's set this up in the form of an equation and solve for "x". It would be set up like this:
2x + 20 = 3x - 30
You see up top that it says 2(x + 10). What you would do is multiply what's in the parenthesis by 2, in order to get 2x + 20. Then put the equal sign and write 3x - 30. Subtract 3x from both sides to get -1x + 20 = -30. Subtract 20 from both sides to get -1x = -50. Divide each side by -1 to isolate the "x". In this case, because you are dividing a negative number by a negative number, your quotient will be positive. -50/-1 is 50. Let's plug in the value as "x" and see if it works. 50 * 2 is 100. 100 + 20 is 120. 50 * 3 is 150. 150 - 30 is 120. 120 = 120. There. x = 50.
Answer:
A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.
The given equation consisting of variable , m is
![\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})](https://tex.z-dn.net/?f=%5Cfrac%7B2%20m%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B2%20m%7D%7B2%20m-3%7D%3D1%5C%5C%5C%5C%202%20m%5B%5Cfrac%7B1%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B1%7D%7B2%20m-3%7D%5D%3D1%5C%5C%5C%5C%202%20m%5Ctimes%20%5Cfrac%7B%5B2%20m-3%20-2%20m-%203%5D%7D%7B4m%5E2-9%7D%3D1%5C%5C%5C%5C%20-6%20%5Ctimes%202%20m%3D4%20m%5E2%20-9%5C%5C%5C%5C%204%20m%5E2%20%2B1%202%20m%20-9%3D0%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%5Csqrt%7B12%5E2-4%20%5Ctimes%204%20%5Ctimes%20%28-9%29%7D%7D%7B2%5Ctimes%204%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B144%2B144%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B288%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%2012%20%5Csqrt%7B2%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B3%7D%7B2%7D%5Ctimes%28-1%20%5Cpm%20%5Csqrt%7B2%7D%29)
None of the two solution
, is extraneous.
Here, L.H.S= R.H.S
Option A: 0→ extraneous
