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3 years ago

Given the points (3, -1) and (-8, 0) find the slope.

Mathematics

1 answer:

Alborosie3 years ago

3 0

Answer:

Step-by-step explanation:

(x₁ , y₁) = (3,-1) &(x₂,y₂) = (-8, 0)

$Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{0-[-1]}{-8-3}\\\\=\frac{0+1}{-11}\\\\=\frac{1}{-11}\\\\=\frac{-1}{11}$

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Need help!!!!!!!!!!!!!!!

icang [17]

Answer:

no error original system so good by

8 0

3 years ago

What is √-11 in the form a+bi?

Mice21 [21]

Given:

The number is $\sqrt{-11}$ .

To find:

The a+bi form of given number.

Solution:

We have,

$\sqrt{-11}$

It can be written as

$\sqrt{-11}=\sqrt{-1\times 11}$

$\sqrt{-11}=\sqrt{-1}\times \sqrt{11}$ $[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]$

$\sqrt{-11}=i\times \sqrt{11}$ $[\because \sqrt{-1}=i]$

$\sqrt{-11}=\sqrt{11}i$

Here, real part is missing. So, it can be taken as 0.

$\sqrt{-11}=0+\sqrt{11}i$

So, a = 0 and $b=\sqrt{11}$ .

Therefore, the a+bi form of given number is $0+\sqrt{11}i$ .

5 0

3 years ago

F(x)

mina [271]

a) Since both limits are distinct and do not exist, we conclude that x = - 1 is not part of the domain of the rational function.

b) The function $f(x) = \frac{x}{x^{2}+ x}$ is equivalent to the function $g(x) = \frac{1}{x + 1}$ .

<h3>How to determine whether a limit exists or not</h3>

According to theory of limits, a function f(x) exists for x = a if and only if $\lim_{x\to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$ . This criterion is commonly used to prove continuity of functions.

Rational functions are not continuous for all value of x, as there are x-values that make denominator equal to 0. Based on the figure given below, we have the following lateral limits:

$\lim_{x \to -1^{-}} \frac{x}{x^{2}+x} = - \infty$