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kodGreya [7K]
3 years ago
7

Determine the horizontal asymptote for r(x) = x^3-2x^2+3 / x-2, if one exists.

Mathematics
2 answers:
Sauron [17]3 years ago
8 0
There is no horizontal asymptote.
Leno4ka [110]3 years ago
4 0

Answer:

Step-by-step explanation:

A horizontal asymptote exists for this function at y = 0.

Compare the degrees of the numerator and denominator

If N < D, then y = 0 is the horizontal asymptote.

If N = D, then y =  

a

b

(a = leading coefficient numerator and b = leading coefficient denominator).

If N > D, then there is no horizontal asymptote.

If N > D by only one, then there is a slant asymptote.

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Please solve from A thru D. Thanks
Mashutka [201]
The unkown side legnth is 9 centimeteres long all you had to do was subtract all the numbers from 44 and what your left with is the unknown side.
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3 years ago
Mindy conducts an experiment with three quarters. She counts the number of heads and tails for each flip of the three quarters.
valkas [14]

Answer:

i don't have the table so no one can answer

Step-by-step explanation:

4 0
2 years ago
You are ordering pencils from Amazon for $3.50 each. Shipping costs an additional $2.99. This situation can be represented with
DENIUS [597]

Answer:

Y=total cost of purchasing the pencils

Step-by-step explanation:

Y=3.5x+2.99

Where,

Y=total cost of purchasing the pencils

3.5=price of each pencil

X= number of pencils bought

2.99= shipping cost

For instance, if you order for 5 pencils on Amazon, the total cost of purchasing it is

Y=$3.50(5)+$2.99

Y=$17.50+$2.99

Y=$20.49

3 0
3 years ago
Last month, Katie started training for the swim team. She swam 20 laps every day for 18 training days. This month, Katie wants t
nasty-shy [4]

She swim 20 laps every day for 18 training days

so total laps done last month = 20*18 = 360 laps

She will swim 3 more laps per training day and add more training days to her monthly schedule.

20 laps + 3 laps = 23 laps

(a)If Katie adds 5 training days to her schedule this month, will she swim 215 more laps?

18 days + 5 days more = 23 days

23 laps every day for 23 training days

so total laps = 23*23 = 529 laps. That is 169 more laps than 360 laps done last month. Hence she cannot swim 215 more laps

If Katie adds 10 more training days to her schedule this month, will she swim 215 more laps?

18 days + 10 days more = 28 days

23 laps every day for 28 training days

so total laps = 23*28 =644 laps. That is 284 more laps than 360 laps done last month. Hence she can swim 215 more laps

(c)If Katie adds 6 more training days to her schedule this month, will she swim 215 more laps?

18 days + 6 days more = 24 days

23 laps every day for 24 training days

so total laps = 23*24 =552 laps. That is 192 more laps than 360 laps done last month. Hence she cannot swim 215 more laps

Build an equation for finding the number of additional training days she needs to swim 215 more laps

Let x be the number of days more

So training days = 18 + x

215 more laps , so total laps = 360 + 215 = 575

3 more laps per training day so 20 + 3 = 23 laps per day

laps * training days = total laps

so equation becomes

23 * (18+x) = 575

Now solve for x

414 + 23x = 575

Subtract 414 from both sides

23x = 161

x= 7

so she needs 7 training days to swim 215 more laps

3 0
3 years ago
What is the slope of a line that is parallel to the line y = 3/4x + 2?
alexandr402 [8]
3/4
when it’s parallel it has the same slope!
5 0
3 years ago
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