All categories

2 years ago

PLSS HELP MATH !!!! 12/11 - ( -2/3 ) = ?

Mathematics

2 answers:

Marysya12 [62]2 years ago

4 0

The answer is 58/33. Or decimal wise, 1.757576

sveta [45]2 years ago

3 0

Answer:

Could this be the answer?

=1^0 ×( 3-1 ) assuming you forgot the close bracket

=1×2

Step-by-step explanation:

You might be interested in

What is the difference? StartFraction 2 x + 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x + 5 Over x cubed minu

inn [45]

Answer:

$\frac{(x+5)(x+2)}{x(x-3)(x+3)}$

Step-by-step explanation:

The given expression is

$\frac{2x+5}{x^{2} -3x} -\frac{3x+5}{x^{3} -9x} -\frac{x+1}{x^{2}-9}$

First, we need to factor each denominator

$\frac{2x+5}{x(x-3)} -\frac{3x+5}{x(x+3)(x-3)} -\frac{x+1}{(x-3)(x+3)}$

So, the least common factor (LCF) is $x(x-3)(x+3)$ , because they are the factors that repeats.

Now, we diviide the LCF by each denominator, to then multiply it by each numerator.

$\frac{(x+3)(2x+5)-3x-5-x(x+1)}{x(x-3)(x+3)} =\frac{2x^{2}+5x+6x+15-3x-5-x^{2}-x }{x(x-3)(x+3)}\\\frac{x^{2}+7x+10}{x(x-3)(x+3)}$

Then, we factor the numerator, to do so, we need to find two numbers which product is 10 and which sum is 7.

$\frac{(x+5)(x+2)}{x(x-3)(x+3)}$

Therefore, the expression is equivalent to

$\frac{(x+5)(x+2)}{x(x-3)(x+3)}$

8 0

2 years ago

Read 2 more answers

Witch of the following numbers is clostest to 6?

Reika [66]

3,2,1,6 are all factors of 6.

6 0

3 years ago

For the data set 100, 100, 104, 107, 107, the mean is 103,. What is the mean absolute deviation

Anna35 [415]

Answer:

Mean Absolute Deviation is 3.

Step-by-step explanation:

103-100=3

103-104=1

103-107=4

3+3+1+4+4=15/ the five numbers

15/5= 3

5 0

3 years ago

What is the value of |C| in this matrix

damaskus [11]

Answer:

-12

Step-by-step explanation:

required value is det(C)

detC = (-6×4)-(6×-2)

so |C| = -24+12

=-12

6 0

3 years ago

Square root of 9/16

Aleksandr-060686 [28]

$\sqrt{ \frac{9}{16} } = \frac{ \sqrt{9} }{ \sqrt{16} } = \frac{3}{4}$

5 0

3 years ago