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Ad libitum [116K]
3 years ago
8

Given the functions a(x) = 4x2 − x + 2 and b(x) = x + 1, identify the oblique asymptote of the function a of x over the function

b of x.

Mathematics
1 answer:
olga2289 [7]3 years ago
5 0
a(x)=4x^2-x+2 \\
b(x)=x+1 \\ \\
f(x)=\frac{a(x)}{b(x)}=\frac{4x^2-x+2}{x+1}

You can find the oblique asymptote by dividing the polynomial in the numerator by the polynomial in the denominator. The asymptote is the result of the division, without the remainder.
Look at the picture.

The oblique asymptote is y=4x-5.

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Hey lil help here i need to find an equation and solve it plz help
marta [7]

29.50x + 548 = 2023

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then devide 29.50 from it's self and the remaining
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3 years ago
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Consider these three numbers written in scientific notation: 6.5 × 103, 5.5 × 105, and 1.1 × 103. Which number is the greatest,
horrorfan [7]

Answer:

6.5×10'3=6,500.

5.5×10'5=550,000

1.1×10'3=1,100

so as u can see the second number is greater

by 548,900

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3 years ago
Not sure of this one. can an.expert help
kifflom [539]

How to get answer by JKismyhusbandbae: To get the answer you would multiply

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6 0
4 years ago
As an estimation we are told £3 is €4.<br> Convert £18 to euros.
Virty [35]

The value of £18 to euros will be €24.

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

Given that:-

As an estimation, we are told £3 is €4. Convert £18 to euros.

The conversion will be done as:-

£3     =    €4

£ 1     =   €4/ 3

18 x £ 1 = ( €4/ 3 )  x 18

£18       =  €24  

Therefore the value of £18 to euros will be €24.

To know more about expression follow

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8 0
2 years ago
4. Parking fees at IIUM are RM 5.00 for IIUM students and RM 7.50 for non-IIUM students. At the
professor190 [17]

Answer:

(a) 780 students and 960 non-students  

(b) No. The maximum revenue is RM9000 from 1200 non-students.

(c). Revenue is maximum of RM9000 at 1200 non-students, decreasing by RM2.50 per student to a minimum of RM6000 at 1200 students

Step-by-step explanation:

 Let x = IIUM students and

and  y = non-IIUM students

You have two conditions

(a)          x +         y = total vehicles parked

(b) 5.00x + 7.50y = total gross receipts

(a) Wednesday

From your table,  

       x +          y = 1740

5.00x + 7.70y = RM11 100  

Solve the simultaneous equations

\begin{array}{rrcrl}(1) & x + y & = &1740&\\(2) & 5.00x + 7.50y & = & 11 100\\(3)&  5.00x + 5.00y & = & 8700 & \text{Multiplied (1) by 5}\\&2.50 y & = &2400 &\text{Subtracted (3) from (2)}\\(4)&y& = &\mathbf{960} &\text{Divided each side by 2.50}\\& x +960& = &1740& \text{Substituted (4) into (1)}\\& x& = &\mathbf{780}& \\\end{array}\\\text{There are $\large \boxed{\textbf{780 students and 960 non-students}}$}

(b) Can 1200 vehicles bring in RM10000?

No. Even if all the cars were from non-students, the most you could get is  

1200 × 7.50 = RM9000

(c) Possible combinations for 1200 vehicles  

Revenue = 5.00x + 7.50y = 5.00x + 7.50(1200 -x) = 5.00x + 9000 - 7.50x =  

Revenue = 9000 - 2.50x

The maximum revenue of RM9000 occurs when there are no student cars and 1200 non-student cars.

For each student car that enters and displaces a non-student, the revenue drops by RM2.50.

Finally. when there are 1200 student cars and no non-students, the revenue has dropped to a minimum of RM6000.  

3 0
3 years ago
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