Mathematics, 22.07.2019 22:30, zavalaadrian846

Ineed with these removable discontinuities.

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answered: myiacoykendall

First problem: a (2,0)

Second problem: b. none of these; the answer is (4, 5/3) which is not listed.

Third problem: b. none of the above; there are no holes period.

Step-by-step explanation:

First problem: The hole is going to make both the bottom and the top zero.

So I start at the bottom first.

x^2-3x+2=0

The left hand expression is factorable.

Since the coefficient of x^2 is 1, you are looking for two numbers that multiply to be 2 and add to be -3.

Those numbers are -2 and -1 since (-2)(-1)=2 and -2+(-1)=-3.

The factored form of the equation is:

(x-2)(x-1)=0 .

This means x-2=0 or x-1=0.

We have to solve both equations here.

x-2=0

Add 2 on both sides:

x=2

x-1=0

Add 1 on both sides:

x=1

Now to determine if x=2 or x=1 is a hole, we have to see if it makes the top 0.

If the top is zero when you replace in 2 for x, then x=2 is a hole.

If the top is zero when you replace in 1 for x, then x=1 is a hole.

Let's do that.

x^2-4x+4

x=2

2^2-4(2)+4

4-8+4

-4+4

So we have a hole at x=2.

x^2-4x+4

x=1

1^2-4(1)+4

1-4+4

-3+4

So x=1 is not a hole, it is a vertical asymptote. We know it is a vertical asymptote instead of a hole because the numerator wasn't 0 when we plugged in the x=1.

So anyways to find the point for which we have the hole, we will cancel out the factor that makes us have 0/0.

So let's factor the denominator now.

Since the coefficient of x^2 is 1, all we have to do is find two numbers that multiply to be 4 and add up to be -4.

Those numbers are -2 and -2 because -2(-2)=4 and -2+(-2)=-4.

$f(x)=\frac{(x-2)(x-2)}{(x-2)(x-1)}=\frac{x-2}{x-1}$

So now let's plug in 2 into the simplified version:

$f(2)=\frac{2-2}{2-1}=\frac{0}{1}=0$ .

So the hole is at x=2 and the point for which the hole is at is (2,0).

a. (2,0)

Problem 2:

So these quadratics are the same kind of the ones before. They all have coefficient of x^2 being 1.

I'm going to start with the factored forms this time:

The factored form of x^2-3x-4 is (x-4)(x+1) because -4(1)=-3 and -4+1=-3.

The factored form of x^2-5x+4 is (x-4)(x-1) because -4(-1)=4 and -4+(-1)=-5.

Look at $\frac{(x-4)(x+1)}{(x-4)(x-1)}$ .

The hole is going to be when you have 0/0.

This happens at x=4 because x-4 is 0 when x=4.

The hole is at x=4.

Let's find the point now. It is (4,something).

So let's cancel out the (x-4)'s now.

$\frac{x+1}{x-1}$

Plug in x=4 to find the corresponding y:

$\frac{4+1}{4-1}{/tex][tex]\frac{5}{3}$

The hole is at (4, 5/3).

Third problem:

x^2-4x+4 has factored form (x-2)(x-2) because (-2)(-2)=4 and -2+(-2)=-4.

x^2-5x+4 has factored form (x-4)(x-1) because (-4)(-1)=4 and -4+(-1)=-5.

There are no common factors on top and bottom. You aren't going to have a hole. There is no value of x that gives you 0/0.

answered: Guest

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step-by-step explanation:

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Write a paragraph proof! ! 20 points and will give brainliest

answered: Guest

solution:

the given function is ,

$f(x) = 4 x^3 - 72 x^2 + 320 x,$

= $4x( x^2- 18 x + 80)\\\\ 4 x(x^2-10 x -8 x + 80) \\\\ 4 x[(x-10)(x-8)]$

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

$f(x) = a x^3+ b x^2 + c x+d,$

sum of the roots = a + b +c= 0+8+10=18= $\frac{-b}{a}$

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 = $\frac{c}{a}$

product of roots = a b c = 0= $\frac{d}{a}$

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function $f(x) = 4 x^3 - 72 x^2 + 320 x,$ is given by entry and exit route of this hall.