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gogolik [260]
3 years ago
15

Odd numbers are always divisible by 2

Mathematics
1 answer:
Stells [14]3 years ago
4 0
Odd numbers are not divisible by 2. Only even numbers are visible by 2. Such as, 4,6,8.

Multiples of 10 do always end in 0.

& i'm not so sure about the last one.

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What binomial do you have to add to the polynomial 2x^2+3y^2-5xy+1 to get the following?
Contact [7]

Binomial -3y^{2}+5xy when added to polynomial 2x^{2} +3y^{2} -5xy+1  gives  polynomial  that does not contain the variable y .

<h3>What is binomial?</h3>

A mathematical expression consisting of two terms connected by a plus sign or minus sign .

Example: x + 2 is a binomial, where x and 2 are two separate terms.

<h3>What is polynomial?</h3>

A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer

According to the question

polynomial 2x^{2} +3y^{2} -5xy+1

when added with  -3y^{2}+5xy  

= 2x^{2}+1   (as 3y^{2} -5xy got cancelled )

that does not contain the variable y

Hence, Binomial -3y^{2}+5xy when added to polynomial 2x^{2} +3y^{2} -5xy+1  gives that does not contain the variable y .

To know more about Binomial and polynomial here :

brainly.com/question/1698358

# SPJ2

3 0
1 year ago
Read 2 more answers
PLEASE HELP I AM BEING TIMED!
german
Answer: C.) Congruent (or third option)
3 0
2 years ago
Define fn : [0,1] --&gt; R by the
sasho [114]

Answer:

The sequence of functions \{x^{n}\}_{n\in \mathbb{N}} converges to the function

f(x)=\begin{cases}0&0\leq x.

Step-by-step explanation:

The limit \lim_{n\to \infty }c^{n} exists and converges to zero whenever \lvert c \rvert. But, if c=1 the sequence \{c^{n}\} is constant and all its terms are equal to 1, then converges to 1. Using this result, consider the sequence of functions \{f_{n}\} defined on the interval [0,1] by f_{n}(x)=x^{n}. Then, for all 0\leq x we have that \lim_{n\to \infty}x^{n}=0. Now, if x=1, then \lim_{n\to \infty }x^{n}=1. Therefore, the limit function of the sequence of functions is

f(x)=\begin{cases}0&0\leq x.

To show that the convergence is not uniform consider 0. For any n>1 choose x\in (0,1)  such that \varepsilon^{1/n}. Then

\varepsilon

This implies that the convergence is not uniform.

8 0
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Aleonysh [2.5K]

Answer:

Step-by-step explanation:

THE ANSWER IS 9

6 0
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A football field is 53 1/3 yards wide how many feet wide is a football field
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