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slavikrds [6]
3 years ago
13

The factors of 12pq+3pr+8qr+2r² are ______________.

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
4 0
I belive its number one :)
Xelga [282]3 years ago
4 0
The first choice would be correct
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A bakery offers a sale price of $3.50 for 4 muffins. What is the price per dozen?
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Read 2 more answers
Graph y = -2 and x= 3 individually state the slope and x and y intercept form for each
attashe74 [19]

y = -2 and x= 3

Graph is attached below. Black line is the graph of y=-2

Blue line is the graph of x=3

Whenever we get equation like x = something, in that case slope is always undefined

Whenever we get equation like y = something, in that case slope is always 0

y = -2 and x= 3

For x=3, the slope is undefined.

The graph of x=3 is a vertical line at 3 on x. The x intercept is 3 and there is no y intercept.

For y=-2, the slope is 0

The graph of y=-2 is a horizontal line at -2 on y. The y intercept is -2 and there is no x intercept.


4 0
3 years ago
Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i
liberstina [14]

Answer:

  ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

  \diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}

8 0
1 year ago
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