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LuckyWell [14K]
3 years ago
10

Nadiya determines that there are no fractions equivalent to StartFraction 2 Over 10 EndFraction with a denominator greater than

10, but less than 20, that have a whole number numerator and denominator. Is Nadiya correct?
Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to StartFraction 2 Over 10 EndFraction is StartFraction 4 Over 20 EndFraction.
Yes, Nadiya is correct. By multiplying the whole fraction by any whole number, each equivalent fraction will continue to have a denominator of 10.
No, Nadiya is incorrect. She could multiply both the numerator and denominator of the unit fraction equivalent to StartFraction 2 Over 10 EndFraction by 3 and arrive at the equivalent fraction StartFraction 3 Over 15 EndFraction.
No, Nadiya is incorrect. She could add any number between 1 and 8 to the numerator and denominator and arrive at an equivalent fraction with a denominator greater than 10 but less than 20.
Mathematics
2 answers:
Mila [183]3 years ago
6 0

Answer:

the first one.

Step-by-step explanation:

2/10=4/20 you cant have any other whole number fraction in between those that is equivalent.

Yanka [14]3 years ago
4 0

Answer:

3 Over 15

Step-by-step explanation:

i got 100.

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Choose the number that belongs to the set described
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Natural numbers,integers, rational numbers, irrational numbers.

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Find the factorization of each of the following number:<br> A.18 B.24 C.38 D.81
const2013 [10]

Answer:

A: (3^2)*2

B: (2^3)*3

C: 2*19

D: (3^2)*9

Step-by-step explanation:

We have:

A) 18= 3*3*2= (3^2)*2

B) 24=2*2*2*3=(2^3)*3

C) 38 = 2*19

D) 81 =3*3*9=(3^2)*9

6 0
2 years ago
What is the coefficient of x2y3 in the expansion of (2x + y)5?
Zigmanuir [339]

Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

(2 x+y)^{5}

Using binomial theorem:

(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}

Here a=2 x, b=y

Substitute in the binomial formula, we get

(2x+y)^5=\sum_{i=0}^{5}\left(\begin{array}{l}5 \\i\end{array}\right)(2 x)^{(5-i)} y^{i}

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.

$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

5 0
3 years ago
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