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kompoz [17]
4 years ago
12

HELP!!!!!!!!!! Determine the horizontal asymptote for the rational function.

Mathematics
2 answers:
Nata [24]4 years ago
7 0

Answer:

y = 1

Step-by-step explanation:

Took test

(Please give me brainiest to help me reach the next level, only a few more brainiest to go!)

Brilliant_brown [7]4 years ago
4 0

Answer:

  y = 1

Step-by-step explanation:

"The graph approaches an imaginary horizontal line at y equals 1" means the horizontal asymptote is y = 1.

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I will give brainlyiest, 5 stars!, a thanks and also a friend request if you get this right
Angelina_Jolie [31]

Answer:

the answer is n=>8

Step-by-step explanation:


5 0
3 years ago
Just some simple fraction problems, I am stuck on this though.
Zina [86]

Answer:

5 1/6 ÷ 4 2/3 = 31/28

3 3/4 ÷ 1 2/8 = 3

4 2/4 ÷ 2 1/4 = 2

5 2/5 ÷ 1 4/5 = 3

4 2/8 ÷ 3 2/4 = 17/4

3 3/9 ÷ 2 4/12 = 10/7

7 1/2 ÷ 3 3/5 = 25/12

6 2/8 ÷ 3 1/3 = 15/8

Explanation:

5 1/6 ÷ 4 2/3 = 31/6 ÷ 14/3 = (31/14) / (6/3) = (31/14) / 2 [when you divide by a certain number you multiply that number by the denominator] = 31/28

5 1/6 ÷ 4 2/3 = 31/6 ÷ 14/3 = 31/6 × 3/14

7 0
3 years ago
(8,5) and (x, -1); slope: -6
Aliun [14]

Answer:

9 = x; (9, -1)

Step-by-step explanation:

-y₁ + y₂\-x₁ + x₂ = m

-5 - 1\-8 + x

-8 + x = 1

+8 + 8

_________

x = 9

The denominator is set to equal <em>one</em> because the numerator already equals <em>negative</em><em> </em><em>six</em><em>,</em><em> </em>and all we have to do is divide it by a POSITIVE <em>one</em><em> </em>to stay at <em>negative</em><em> </em><em>six</em><em>.</em>

I am joyous to assist you anytime.

3 0
4 years ago
What is the answer please.... need help
mina [271]

Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

as

Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

8 0
3 years ago
C. Find the missing numbers in the number sentences below.
madreJ [45]

Answer:

7 x 1 = 8 x 7 I think

Step-by-step explanation:

so I hope it helps

3 0
2 years ago
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