All categories

3 years ago

0.11 as a fraction and in the simplest form

Mathematics

2 answers:

weeeeeb [17]3 years ago

6 0

The simplest form it can be put into is 11/100 but the approximate value is 10.91 or even closer is 1/9

vovangra [49]3 years ago

5 0

Considering that the last 1 in 0.11 is located in the hundredths place, the simplest from of this fraction is 11/100 because it cannot be reduced.

*If you have any questions, feel free to ask me! I'd be happy to help you anytime!

You might be interested in

Darren is finding the equation in the form y = m x b for a trend line that passes through the points (2, 18) and (–3, 8). Which

Lera25 [3.4K]

B is consider to be the y-intercept. If you plot those two points and draw a line connecting the two. The line would go through (0,14). Making 14 the y-intercept and the correct answer.

5 0

2 years ago

Perpendicular Vectors:

klemol [59]

1.
v x w = 8 i - 24 j + 15 k + 10 j - 12 i - 24 k =
= - 4 i - 14 j - 9 k
Answer: D )
2.
v x w = -16 i - 16 j - 18 k + 12 j + 48 i - 8 k =
= 32 i - 4 j - 10 k
Answer: C )
3.
The cross product:
< - 6, 7, 2 > x < 8, 5, -3 > =
= - 21 i + 16 j - 30 k - 18 j - 10 i - 56 k =
= - 31 j - 2 j - 86 k = < - 31, - 2, - 86 >
Vectors are perpendicular if: cos ( u, v ) = 0
< - 6 , 7. 2 > * < -31, - 2, - 86 > = 186 - 14 - 172 = 0
< 8, 5 , - 3 > * < - 31, - 2, - 86 > = -248 - 10 + 258 = 0
Answer: A ) < - 31, - 2 , - 86 >, yes.

3 0

3 years ago

I literally need help with this question-<br><br> It’s ASA congruence btw

ryzh [129]

Answer:

AB and DE

Step-by-step explanation:

Look closely and you can notice whats the difference.

4 0

3 years ago

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.