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3 years ago

Why does a horizontal line have a slope of zero, but a vertical line have an undefined slope?

Mathematics

1 answer:

shusha [124]3 years ago

8 0

Refer to the diagram shown below.
Define two positions on each line as A (x₁, y₁) and B(x₂, y₂).
Then the slope is
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Horizontal line:
Define A (0, 1) and B (1,1).
The slope is
$m= \frac{1-1}{1-0}=0$

Vertical line:
Define A( (1,0) and B(1,1).
The slope is
$m= \frac{1-0}{1-1}= \frac{1}{0}$ (undefined)

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Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

3 years ago

Can someone help me out?

Andreyy89

The answer would be around 4.5 or 4.6

5 0

2 years ago

Read 2 more answers

What is -15x + 80 > 170

Korolek [52]

Answer:

Step-by-step explanation:

Value of x should be less than (-6).

x = { .......,-9,-8,-7}

For example:

-15*-7 +80 = 105+80 = 185 >170

3 0

2 years ago

What is the order of operation of 100 + 6² - 9

Kipish [7]

Okay so you would use PEMDAS (parentheses, exponents, multiply, divide, add, and subtract) so there are no parentheses so you do the exponents
${6}^{2} = 36$
then you do the rest so 100+36=136 and then 136-9=127 so the answer is 127

6 0

2 years ago

Can somebody help me please!!!

Effectus [21]

Answer:

Part A: 0

Part B: 0 or 1

Step-by-step explanation:

Part A: anything to the 0 power is 1 (x^0= x^y-y) if you solve 7^2 you get 49. 49^0= 1

Part B: Because it is already being raised to the power of 0, if x=0 it will remain equal to one (as explained above). Anything raised to the 1 power is itself, so 7^0= 1 and 1^1 =1

Hope this helped!

3 0

3 years ago