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True [87]
3 years ago
5

What do is 0 divided by0

Mathematics
2 answers:
AURORKA [14]3 years ago
4 0
You can not divide by zero in math.
IrinaVladis [17]3 years ago
3 0
You can't divide by 0 in math, you get 0
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Helping hearts is making care packages. 36 bars of soap and 27 tubes of toothpaste our available each package will contain an eq
AfilCa [17]
The answer is 27 because there isn't enough bars of soap to have 2 in each package
6 0
3 years ago
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Jamie has 7/10 of a pound of chicken to cook for dinner if she's going to put the chicken in between three people how much chick
AURORKA [14]

Answer:

Each person will get \frac{7}{30} or approximately 0.23 pounds of chicken.

Step-by-step explanation:

This is a word problem involving fractions.  Using key words and phrases from the problem, we can decide which operation we need to use to solve.  Since Jamie is splitting the chicken equally between three people, she would need to divide the amount of chicken she has by 3:

\frac{7}{10}÷\frac{3}{1}

Since we can't divide fractions, we need to use the 'keep-change-flip' rule to keep the first fraction, change the operation to multiplication and flip the second fraction:

\frac{7}{10}*\frac{1}{3}=\frac{7}{30}

We can convert the fraction into a decimal by dividing the numerator by the denominator:  7÷30≈0.23 pounds per person.

5 0
4 years ago
What is the simplified expression for 5ab+9ab-ab
Ulleksa [173]
The simplified expression is 13ab
3 0
3 years ago
Find the zeros (roots) of the following equations. f(x) = 2x5 - 9x4 + 12x3 - 12x2 + 10x - 3 = 0
Talja [164]
Hi,
f(1)=2*1^5-9*1^4+12*1^3-12*1^2+10*1-3\\
=2-9+12-12+10-3\\
=0\\

f(x)=(x-1)(2x^4-7x^3+5x^2-7x+3)\\


g(x)=2x^4-7x^3+5x^2-7x+3\\

g(3)=2*3^4-7*3^3+5*3^2-7*3+3\\
=162-189+45-21+3\\
=0\\

g(x)=(x-3)(2x^3-x^2+2x-1)\\

h(x)=2x^3-x^2+2x-1\\

h( \dfrac{1}{2} )=2* \dfrac{1}{8} - \dfrac{1} {4}+1-1\\
=0\\

h(x)=(2x-1)(x^2+1)\\


f(x)=(x-1)(x-3)(2x-1)(x+i)(x-i)\\



Roots are 1,3,0.5,i,-i.

8 0
3 years ago
Choose whether it's always, sometimes, never 
Keith_Richards [23]

Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.

Explanation:

1)

The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.

If a\in Z\text{ and }b\in Z, then a+b\in Z.

Therefore, an integer added to an integer is an integer, this statement is always true.

2)

A polynomial is in the form of,

p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0

Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we subtract the two polynomial then the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3)

If a polynomial divided by a polynomial  then it may or may not be a polynomial.

If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{f(x)}{g(x)}=x^2+5, which a polynomial.

If the degree of numerator polynomial is less than the degree of denominator polynomial then it is a rational function.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{g(x)}{f(x)}=\frac{1}{x^2+5}, which a not a polynomial.

Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.

4)

As we know a polynomial is in the form of,

p(x)=a_nx^n+a_{n-1}x^{x-1}+...+a_1x+a_0

Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3 0
3 years ago
Read 2 more answers
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