All categories

3 years ago

WILL MARK BRAINLIEST PLEASE HELP

Mathematics

1 answer:

vlada-n [284]3 years ago

6 0

Answer:

2x + (-2x + 3) = 3

2x - 2x +3

0 + 3

2x-7 + (-2x) = -7

2x -2x - 7

0 -7

-7

-(5x -1) + 5x = 1

-5x +1 + 5x

-5x + 5x + 1

0 + 1

Step-by-step explanation: In addition, the parentheses help identify the individual addends. If adding a negative number, it is the same as subtracting. If there is a negative sign outside the parentheses, as in example c, you must distribute it: the signs of the numbers within the parentheses must be reversed when the parentheses are removed.

You might be interested in

Fiona bought some socks that cost $4.95 for each pair and some belts that cost $6.55 each. Fiona spent $27.95 in all. Let a repr

Nookie1986 [14]

Answer:

4.95a + 6.55b = 27.95

3 0

3 years ago

Read 2 more answers

Add the fractions simplify your answer 2 7/8 + 2 3/8

Mashutka [201]

Answer:

5 1/4 or 21/4

7 0

3 years ago

Read 2 more answers

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

Holly missed 6 of 181 school days this year. Approximately what percent of school days was Holly in attendance?

Natasha2012 [34]

$181-6=175\\\\\frac{175}{181}\cdot100\%=\\\\\frac{17500}{181}\%=\\\\96.7\%\approx97\%\\\\Answer\ C$

4 0

3 years ago

Whice set of side will make a triangle

telo118 [61]

First things first.. What type of triangle

If u are taking about a normal triangle here is some:

∠60+∠60+∠60

∠30+∠90+∠60

∠11+∠82+∠87

7 0

3 years ago