All categories

3 years ago

-2 2 the second power

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

3 0

-4 or -484
if it’s -2 to the 2nd power then it is -4
or if it’s -22 to the 2nd power then it is -484

You might be interested in

What is the midpoint

antoniya [11.8K]

Answer:

(0.5, -0.5)

Step-by-step explanation:

We are given a line segment on the graph with two known points (ending points) and we are to find its mid point.

We know the formula for the mid point:

$Midpoint = ( \frac {x_1 + x_2} {2} ,\frac {y_1 + y_2} {2} )$

Substituting the coordinates of the given points in the above formula:

Mid point = $(\frac {5 + (-4) } {2} , \frac { 6 + (-7)} {2} )$ = (0.5, -0.5)

8 0

3 years ago

Read 2 more answers

Takeisha is placing placemats around a table. Each placemat is 14 inches long and 10 inches wide. The rectangular table is 42 in

Vesnalui [34]

welll i kinda don get ur question but ummmm if u say like the graetest number then 48

8 0

3 years ago

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

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

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

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

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

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

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

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

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$

Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

$\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}$

Converting ÷ sign into multiplication we will take reciprocal of the second term

$=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}$

Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.

5 0

3 years ago

13 is 29% of what number?

Zielflug [23.3K]

13/23% = 1300/23 (56.52)

Therefore, 13 is 29% of 1300/23 (56.52)

4 0

3 years ago

I am a number between 7 000 000 and

Anon25 [30]

First, start off by listing a few of the numbers that follow the first three clues to see if you can narrow it down.
7,111,111
7,333,333
7,555,555
etc.
Then, start adding up digits to see if you're getting close.
7+1+1+1+1+1+1 = 13
7+3+3+3+3+3+3 = 25
7+5+5+5+5+5+5 = 37
Since 7,555,555 is too high, we step it down to 7,555,333
7+5+5+5+3+3+3 = 31
7,555,333 will work as an answer, as well as 7,333,555, since it's the same amount when the digits are added together.

4 0

3 years ago

-2 2 the second power​

-2 2 the second power