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3 years ago

Please help me I give points and have a a explication

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

7 0

Answer:on all of them?

Step-by-step explanation:

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What is the slope of the line?

Eddi Din [679]

Answer:

-1

Step-by-step explanation:

Choose two point on that line such as (1,1) and (2,0)

Then use slope formula y-y/x-x

1-0/1-2 = 1/-1 = -1

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3 years ago

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-2(2n-2)>18-3n how do you solve this inequality

Svet_ta [14]

-2 * 2n -2 equals -4n + 4 > 18n -3n

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3 years ago

What is tje equation for the linear function whose graph contains the points (2,5) and (3,0) ?

dolphi86 [110]

In order to find a slope, there is an equation:

Y2-Y1/X2-X1

X1=2
X2=3
Y1=5
Y2=0

0-5= -5
3-2=1

Since slope is written as Y/X, slope would be-5/1 or simply, -5.

Now, we need to find the y-intercept.

To do this, shift the original equation:

Original equation: y=-5x+b

Shifted equation: b=y+5x

Now, plug in one of the points.

Let's use (2,5)

b=5+5(2)

5*2=10
10+5=15

y-intercept=15

So, the full equation of the line would be:

y=-5x+15

4 0

4 years ago

Pls help this is pretty urgent

KIM [24]

Answer:

(a) 0

(b) f(x) = g(x)

Step-by-step explanation:

Given rational function:

$f(x)=\dfrac{x^2+2x+1}{x^2-1}$

Part (a)

Factor the numerator and denominator of the given rational function:

$\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}$

Substitute x = -1 to find the limit:

$\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0$

Therefore:

$\displaystyle \lim_{x \to -1}f(x)=0$

Part (b)

From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).

Part (c)

As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.

$\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty$

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Answer:

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