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serg [7]
3 years ago
10

Limit x->1 x*2 - 1 / (x-1)*2

Mathematics
1 answer:
hoa [83]3 years ago
4 0
\rm \lim\limits_{x\to1}\dfrac{x^2-1}{(x-1)^2}\quad=\quad\lim\limits_{x\to1}\dfrac{(x-1)(x+1)}{(x-1)(x-1)}

The numerator factors into conjugates
while we can also expand the square in the denominator.
Apply a cancellation,

\rm \lim\limits_{x\to1}\dfrac{x+1}{x-1}

When you started, you had a function which was approaching this indeterminate form 0/0. But now, the numerator is approaching 2 while the denominator approaches 0.

This is behaving exactly like this limit,

\rm \lim\limits_{x\to0}\dfrac1x

The bottom is getting really small,
so the fraction overall is getting huge, approaching infinity.

Oh actually with your problem,
from the right side of 1, the denominator is positive,
so we blow up to infinity.

but from the left side of 1, the denominator is negative,
so we blow up to negative infinity.

So the limit does not exist if it's just x->1.
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How many different 7-digit telephone numbers are possible if the first digit cannot be 0, but the last two digits must be 0?
solniwko [45]
The first digit can only range from 1-9, resulting in 9 possible options. The next four digits can range from 0-9, resulting in 10 options for each. Since the last two digits remain the same, they do not affect the sample size. Using the fundamental counting principle, we can find the amount of telephone numbers possible in the following equation.
9 \times 10 \times 10 \times 10 \times 10 \times 1 \times 1 \\ 90000

4 0
3 years ago
2+x/6 - x+6/10 = <br> Step by step explanation pls
Nikitich [7]

click on it to view full solution.

I think it is right

4 0
3 years ago
Select the correct answer for each statement.
Alex787 [66]
<span>
1.) Will yield consecutive odd integers.

k+10, k+12, k+14
or
k+2, k+3, k+4

The answer is k + 10, k +12 , k + 14

Because consecutive integers have difference of 2.

If k is odd, then k+10, k+12, and k+14 are consecutive odd integers.

For example, assume k = 1, then,

k+10=11
k+12=13
k+14=15

And 11, 13 and 15 are consecutive odd integers.

2.) Will yield consecutive integers.

k+1, k+2, k+3
or
k+6, k+8, k+10

The answer is k+1, k+2, k+3

Let k be any integer number, k+1, k+2, k+3 are consecutive integers.

For example, let k = 23

k+1=24
k+2=25
k+3=26

24,25, and 26 are consecutive integers.
</span>
3 0
4 years ago
18. Find the center, vertices, and foci of the ellipse with equation x squared divided by 225 plus y squared divided by 625 = 1.
KengaRu [80]

Answer:

Center (0,0)

Vertices (-15,0), (15,0), (0,-25), (0,25)

Foce (0,-20), (0,20)

Step-by-step explanation:

You are given the ellipse equation

\dfrac{x^2}{225}+\dfrac{y^2}{625}=1

The canonical equation of ellipse with center at (0,0) is

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

So,

a^2=225\Rightarrow a=15\\ \\b^2=625\Rightarrow b=25

Hence, the center of your ellipse is at (0,0) and the vertices are at points (-15,0), (15,0), (0,-25) and (0,25)

This ellipse is strengthen in y-axis, so

c=\sqrt{b^2-a^2}=\sqrt{625-225}=\sqrt{400}=20

and the foci are at points (0,-20) and (0,20).

6 0
3 years ago
Write the equation in slope-intercept form of the line below
lyudmila [28]

Answer:

y = 2

Step-by-step explanation:

6 0
3 years ago
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