All categories

3 years ago

A____ function is a function that can be expressed as the quotient of two polynomial functions

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

7 0

Answer:

rational function

Step-by-step explanation:

That would be a "rational function;" it's the "ratio" of two polynomial functions, with the understanding that the denominator may not be zero.

r-ruslan [8.4K]3 years ago

6 0

Answer:

Rational expression

Step-by-step explanation:

A rational expression is a quotient of two polynomials, where the polynomial in the denominator is not zero. Rational expressions can often be simplified by removing terms that can be factored out of the numerator and denominator. These can be either numbers or functions of x

You might be interested in

The second statement is the of the first.

kipiarov [429]

B is the answer b.converse

5 0

2 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

How do you prove a rectangle in geometry?

dedylja [7]

- The opposite sides are parallel and congruent. - The diagonals bisect each other.

- There are 4 right angles. - The diagonals are congruent.

- Show that both pairs of opposite sides are congruent.

6 0

3 years ago

Y=-x+7

Allisa [31]

Question 1:
y=-x+7
y=2x-2

(-x+7)=2x-2
-x+7+x=2x-2+x
7=2x-2+x
7+2=2x-2+x+2
9=3x
9/3=3x/3
3=x

y=-x+7
y=-(3)+7
y=4

Question 2:
y-1=2x
y+2x=5

y-1=2x
y-1+1=2x+1
y=2x

(2x+1)+2x=5
2x+1+2x=5
4x+1=5
4x+1-1=5-1
4x=4
4x/4=4/4
x=1

y-1=2(1)
y-1+1=2+1
y=3

Answer: x=1 y=3

6 0

2 years ago

Can someone help 6-11x=7x-12

Luden [163]

6-11x = 7x-12

6+12 = 11x+7x

18 = 18x

x = 1

6 0

3 years ago

Read 2 more answers