-2x^(3) +3=-1x^(3) -1x^(2) +1x+1

1 answer:

7 0

-2x³ + 3 = -x³ - x² + x + 1

Add x³ to both sides, and subtract 3 from both sides.

-2x³ + x³ = -x² + x + 1 - 3

Add like terms.

-x³ = -x²+ x - 2

Move all of the (X's) to both sides, by adding x² to both sides, and subtracting x form both sides.

-x³ + x² - x = -2

~Hope I helped!~

You might be interested in

HELP FAST PLEASE HURRY I NEED IT TO BE RIGHT

drek231 [11]

Answer:

{x + 4y = 8

{5x + 5y = 10

Step-by-step explanation:

Just add everything like it said, and you will arrive at your answer.

I am joyous to assist you anytime.

5 0

2 years ago

Tristan has found a good job as a bookkeeping clerk after finishing his associates degree at the local community college. He is

Soloha48 [4]

The increased income is affecting Tristan's financial decisions as now he is spending more as well as saving more.

The correct answer is option B.

Tristan is earning quite a bit more in his full time job. He is spending more on rent, utilities, food, and entertainment, as well as saving $350 each month.

8 0

2 years ago

Read 2 more answers

USE NUMBER TWO USE NUMBER TWO

n200080 [17]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Integration of sin^2 2x cos^3 2x dx

lilavasa [31]

$\int\left[\sin^2(2x)\cos^3(2x)\right]dx=\int\left[\sin^2(2x)\cos^2(2x)\cos(2x)\right]dx\\\\\text{Use the trigonometric identity}\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\=\int\left\{\sin^2(2x)[1-\sin^2(2x)]\cos(2x)\right\}dx\\\\\text{substitute}\ \sin2x=t\ \to\ 2\cos2x\ dx=dt\ \to\ \cos2x\ dx=\dfrac{1}{2}\ dt\\\\=\dfrac{1}{2}\int\left[t^2(1-t^2)\right]dt=\dfrac{1}{2}\int(t^2-t^4)dt=\dfrac{1}{2}\left(\dfrac{1}{3}t^3-\dfrac{1}{5}t^5\right)+C$

$=\dfrac{1}{2}\left(\dfrac{1}{3}\sin^3(2x)-\dfrac{1}{5}\sin^5(2x)\right)+C$

4 0

2 years ago

QUESTION 1

Alexeev081 [22]

Answer:

1. 9

2. 10

Step-by-step explanation:

idk if those first 2 are right

6 0

2 years ago