There were 47 ducks swimming in a pond. A dog jumped into the pond and scared 29 of the ducks away. After

Mathematics

1 answer:

d1i1m1o1n [39]2 years ago

4 0

Answer:

Step-by-step explanation:

Originally, there were 47 ducks. After the dog jumped in, there were 29. This means there were 18 ducks.

47 - 29 = 18

After the dog got out 5 groups of 3 ducks came. This means that 15 ducks came back.

5 × 3 = 15

To find the total amount of ducks, add the ducks together.

18 + 15 = 33

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Which represents the polynomial written in standard form?

pshichka [43]

Answer:

Option A.

Step-by-step explanation:

In order to write any polynomial in standard form, we need to check the degree of each term, then write each term in order of degree, from highest to lowest, left to right.

The given polynomial is

$8x^2y^2-\dfrac{3x^3y}{2}+4x^4-7xy^3$

Here, the combine degree of x and y in each term is 4.

If we arrange the terms according to the degree of x, then

$4x^4-\dfrac{3x^3y}{2}+8x^2y^2-7xy^3$

If we arrange the terms according to the degree of y, then

$-7xy^3+8x^2y^2-\dfrac{3x^3y}{2}+4x^4$

Hence, the correct option is A.

3 0

2 years ago

Read 2 more answers

What is the binomial expansion of (x + 2)4? x4 + 4x3 + 6x2 + 4x + 1 8x3 + 24x2 + 32x x4 + 8x3 + 24x2 + 32x + 16 2x4 + 8x3 + 12x2

Margaret [11]

<u>Answer-</u>

$\boxed{\boxed{(x+2)^4=x^4+8x^3+24x^2+32x+16}}$

<u>Solution-</u>

Given expression is $(x+2)^4$

Applying Binomial Theorem

$\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

Here,

a = x, b = 2 and n = 4

So,

$\left(x+2\right)^4=\sum _{i=0}^4\binom{4}{i}x^{\left(4-i\right)}\cdot \:2^i$

Expanding the summation

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

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