Describe and correct the error in finding the solution of the equation 2x=6x

1 answer:

6 0

The error is that both sides of the equation have a variable. So if you were to combine your like terms, then you'd leave one side without a number and if your teacher is saying there is a left over number then it will be 0

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Select either relation (if the set is a relation but not a function), function (if the set is both a relation and a function), o

gavmur [86]

Answer:

B. Function.

Step-by-step explanation:

We have been given a set. We are asked to choose the correct option about the given set.

We know that a set represents a relation, when each x value corresponds to a y-value. In a relation one x-coordinate can have more than one y-values.

We know that a set represents a function, where each x-value corresponds to exactly one y-value.

Two different value can have same y-value to be a function.

We can see that each value of x corresponds to exactly one y-value. The value of y is 2 for all value of x, therefore, all points lie on a horizontal line $y=2$ .

Therefore, our given set a function.

5 0

3 years ago

Read 2 more answers

Determine the given pair of lines of parallel, perpendicular, or intersecting.

podryga [215]

$k:\ y=m_1\cdot x+b_1\ \ \ \ and\ \ \ \ l:\ y=m_2\cdot x+b_2 \\\\k\ ||\ l\ \ \ \Leftrightarrow\ \ \ m_1=m_2\ \ \ \ \ \ and \ \ \ \ \ \ k\ \bot\ l\ \ \ \Leftrightarrow\ \ \ \ m_1\cdot m_2=-1\\----------------------------\\1.\\ k:\ y= \frac{7}{3} x-8\ \ \ and\ \ \ l:\ y=\frac{7}{3} x+3\\\\m_1=m_2=\frac{7}{3}\ \ \ \Rightarrow\ \ \ k\ ||\ l$

$2.\\k:\ y=2x-1\ \ and\ \ l:\ -x+2y=6\ \Rightarrow\ \ 2y=x+6\ \ \Rightarrow\ \ y= \frac{1}{2}x+3 \\\\m_1 \neq m_2\ \ \ and\ \ \ m_1\cdot m_2 \neq -1\ \ \ \Rightarrow\ \ \ the\ lines\ intersect\\\\3.\\k:\ y= \frac{5}{2} x-17\ \ \ and\ \ \ l:\ y=- \frac{5}{2} x-1\ \ \ \Rightarrow\ \ \ the\ lines\ intersect\\\\4.\\k:\ y= \frac{5}{2} x-4\ \ and\ \ l:\ -5x+2y=6\ \Rightarrow \ 2y=5x+6\ \Rightarrow \ y= \frac{5}{2}x +3\\\\m_1=m_2= \frac{5}{2} \ \ \ \Rightarrow\ \ \ k\ ||\ l$

$5.\\k:\ y=- \frac{2}{3} x+12\ \ \ and\ \ l:\ y= \frac{3}{2} x+3\\\\m_1\cdot m_2=- \frac{2}{3}\cdot \frac{3}{2} =-1\ \ \ \Rightarrow\ \ \ k\ ||\ l\\\\6.\\k:\ y= \frac{1}{3}x-13\ \ \ and\ \ \ l:\ y=3x+2\ \ \ \Rightarrow\ \ \ the\ lines\ intersect \\\\7.\\k:\ y=-4x+5\ \ ,\ \ l: -x+4y=-16\ \Rightarrow\ 4y=x-16\ \Rightarrow\ y= \frac{1}{4}x-4 \\\\m_1\cdot m_2=-4\cdot \frac{1}{4} =-1\ \ \ \Rightarrow\ \ \ k\ \bot\ l$

$8.\\k:\ y= \frac{2}{3} x+3 \ ,\ \ l:\ -2x+3y=21 \ \Rightarrow\ \ 3y=2x+21\ \Rightarrow\ y= \frac{2}{3} x+7\\\\m_1=m_2= \frac{2}{3} \ \ \ \Rightarrow\ \ \ k\ ||\ l$

8 0

2 years ago

-4/9f = -3<br><br>Solve for F

mars1129 [50]

Answer:

Step-by-step explanation:

in the picture