Ask question

Ask question

All categories

FromTheMoon [43]

3 years ago

6

escribe falso o verdadero según corresponda: Para comparar las fracciones

7D" id="TexFormula1" title="\frac{7}{13}" alt="\frac{7}{13}" align="absmiddle" class="latex-formula"> y $\frac{2}{11}$ se comparan los denominadores 2 copia y escribe , segun corresponda $\frac{4}{11}$ - $\frac{7}{3}$

1 answer:

Svetlanka [38]3 years ago

4 0

Answer:

sorry bro don't read Spanish but it's not over 3

Step-by-step explanation:

You might be interested in

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

4 0

3 years ago

Solve. *25 points*<br><br> 1/2 divided by 2/3 =?

Colt1911 [192]

Answer:

The answer is 3/4

Step-by-step explanation:

When you divide two fractions, such as 1/2 ÷ 2/3, you have to flip the second fraction and then you simply multiply the numerators with each other and the denominator with each other.

7 0

3 years ago

Dan grew 22 pumpkins but tge rabbits ate 25 pumpkins.how many pumpkins does dan have left

rodikova [14]

Answer:

0

Step-by-step explanation:

0 because he grew 22 but the rabbit ate 25 which does not make sense

8 0

3 years ago

Read 2 more answers

:D hiiii im back with another most likely easy questionnnnnnnnnnnnnnnnnnnnnn

mihalych1998 [28]

Answer: The answer is 75

20/4=5

5*15=75

5 0

3 years ago

Read 2 more answers

10(n^2+n) -6 (n^2 - 2). simplify the following

Setler [38]

Answer:

4n^2 + 22n

Step-by-step explanation:

10(n^2+n) -6 (n^2 - 2)

10n^2 + 10n - 6n^2 + 12

4n^2 + 22n

3 0

3 years ago