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Elenna [48]
3 years ago
10

Can you help me with this please..

Mathematics
1 answer:
jonny [76]3 years ago
8 0

Answer:

0.5

Step-by-step explanation:

Ok, so it's asking for what (1/(x-1) - 2/(x^2-1)) approaches as x approaches 1. Before we deal with the limit, let's simplify the inside.

We want to combine the two fractions into one fraction. Therefore, we need a common denominator.

1/(x-1) is equal to (x+1)/((x+1)(x-1) is equal to (x+1)/(x^2-1).

the inside expression is therefore (x+1)/(x^2-1) - 2/(x^2-1)

which simplifies to (x-1)/(x^2-1).

and that simplifies further to 1/(x+1).

Now this is a continuous function when x = 1, so to find the limit as x approaches 1 of this function, we can by definition just plug 1 in.

limx->1 (1/(x+1)) = 1/2.

The reason why we didn't just plug 1 in at the beginning is because the function wasn't continuous when x was 1.

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Step-by-step explanation:

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Pre-Calculus - Systems of Equations with 3 Variables please show work/steps
inessss [21]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

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3 years ago
What is the sum of the first 14 terms of the sequence an = 6n -12
mina [271]
When n = 1  first term  = -6 
n = 2  second term =  0
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n = 4 4th term = 12

so we have an Arithmetic sequence  first term = -6 and common difference = 6

Sum 14 terms  =  (14/2)[2*-6 + (14-1)*6]

 =  462  answer
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3 years ago
09 CP
Veseljchak [2.6K]

Answer:

the answer should be as follows:

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2 years ago
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AVprozaik [17]
A because the formula is s^2h/3 and if you plug it in, it's A.
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