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Anit [1.1K]
2 years ago
7

Which of the following are incorrect expressions for slope?

Mathematics
2 answers:
vfiekz [6]2 years ago
8 0

Answer:

Option B and C are correct.

\frac{x_2-x_1}{y_2-y_1}

\frac{run}{rise} are the expression incorrect for slope

Step-by-step explanation:

Slope is defined as the change in the dependent variable  relative to the change in the dependent variable

or the ratio of the horizontal changes to vertical changes between any two points on the graph of the line.

The vertical changes between any two points is rise

The horizontal changes between any two points is run.

Formula for slope is given by:

For any two points (x_1, y_1) and (x_2, y_2)

then slope is:

\text{Slope} =\frac{rise}{run}= \frac{y_2-y_1}{x_2-x_1}

or we can write this as:

Δy = y_2-y_1

Δx = x_2-x_1

⇒\text{Slope} = \frac{\triangle y}{\triangle x}

Therefore, the expression which are incorrect for slope  are;

\frac{x_2-x_1}{y_2-y_1}

\frac{run}{rise}

yaroslaw [1]2 years ago
7 0

Answer:

Choice B and choice C are incorrect expression for slope.

Step-by-step explanation:

Slope is defined as the steepness of a line. It is also called gradient.It is denoted by m.

It is ratio of change in the dependent variable and independent variable.

hence, m = Δy / Δx = y₂-y₁ / x₂-x₁

The change in dependent variable is called rise and the change in dependent variable is called run.

So,   m = rise / run

hence, choice B. and choice C. are incorrect expressions for slope.

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Step-by-step explanation:

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1 year ago
I need help please help me thanks please
Vitek1552 [10]

so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation

~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})

\stackrel{slope}{m}\implies\cfrac{\stackrel{rise}{\stackrel{y_2}{15}-\stackrel{y1}{3}}}{\underset{run}{\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{12}{3}\implies 4\\\\\\% point-slope intercept\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}{3}=\stackrel{m}{4}(x-\stackrel{x_1}{1})\\\\\\y-3=4x-4\implies y = 4x-1\implies \blacktriangleright  y = 4x+(-1)\blacktriangleleft

4 0
3 years ago
What is the measure of angle 1?
tatiyna

Answer:

27 degrees

Step-by-step explanation:

The easy way is by remembering the formula (a-b)/2=c, where a is the larger angle, and b is the smallest angle. (90-36)/2=27.

The longer, more drawn out answer goes as follows. See the image to understand the notation I use:

  1. AOE + BOD + BOA + DOE = 360
  2. AOE + BOD = 90 + 36 = 126
  3. BOA + DOE = 360 - AOE - BOD = 234
  4. Since the sum of a triangle's angles is 180, ODE = (180 - DOE) / 2
  5. Likewise, OBA = (180 - BOA) / 2
  6. Since CDE is 180, CDO = 180 - ODE = 180 - (180 - DOE) / 2
  7. Likewise, CBA is 180, so CBO = 180 - OBA = 180 - (180 - BOA) / 2
  8. The interior angles of the irregular polygon CBOD add up to 360, so CBO + CDO + BOD + BCD = 360.
  9. Substituting what we already found, 180 - (180 - BOA)/2 + 180 - (180 - DOE)/2 + 36 + BCD = 360
  10. Cleaning it all up, we get 180 + (BOA + DOE)/2 + 36 + BCD = 360
  11. As we found in line 3, BOA + DOE = 234, so substituting that in, 180 + 117 + 36 + BCD = 360
  12. Finally, solving for BCD (360 - 36 - 117 - 180) we get our answer, 27

Note: The long drawn out method shown above is a way to derive the formula for the secant theorem. You do not need to use this method every time. Just remember, large angle minus small angle, all divided by 2. That is it.

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