(4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y)

11x²y³ - 4xy²

Step-by-step explanation:

Distribute the first parenthesis by 1 and the second by - 1

Given

(4x²y³ + 2xy² - 2y) - (-7x²y³ +6xy² - 2y)

= 4x²y³ + 2xy² - 2y + 7x²y³ - 6xy² + 2y ← collect like terms

= (4x²y³ + 7x²y³) + (2xy² - 6xy²) + (- 2y + 2y)

= 11x²y³ + (- 4xy²) + 0

= 11x²y³ - 4xy²

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !