All categories

3 years ago

Divide 2x^4+3x^3-5x+6 by x^2+3x-2 .

Mathematics

1 answer:

Anni [7]3 years ago

4 0

2x^4 + 3x^3 - 5x + 6 ÷ x^2 + 3x - 2

= 2x^6 + 3x^5 − 2x^3 − 2x^2 + 6/ x^2

You might be interested in

What is a least common factor for 8 and 20?

ludmilkaskok [199]

8=2*2*2
20=2*2*5

lcf=2*2=4

7 0

4 years ago

what 1+2 eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

Gnesinka [82]

Answer:

Step-by-step explanation:

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

6 0

4 years ago

HELP! please explain how to solve the point of maximum growth rate for the logistic function?!?? I'm really confused!

lora16 [44]

Answer:

18 units over 210 units of time

Step-by-step explanation:

the function reaches its limit at 116 and starts at 0 at -94

|-94|+|116|=210

8 0

3 years ago

There are moles in 1.26 × 1024 particles.

EastWind [94]

Answer:

1.26 x 10^(24)

Step-by-step explanation:

hope this helped

6 0

3 years ago

Verify by direct substitution that the given power series is a solution of the indicated differential equation. [Hint: For a pow

shepuryov [24]

Answer:

The given power series $y =\sum^{\infty}_{n=0} {(-1)^n x^{2n}}$ is a solution of the differential equation (1+x^2)y' + 2xy = 0

Step-by-step explanation:

This is a very trivial exercise, follow the steps below for the solution:

Step 1: Since n = 0, 1, 2, 3, 4, ........, Substitute the values of n into equation (1) below.

$y =\sum^{\infty}_{n=0} {(-1)^n x^{2n}}$ .....................(1)

$y = 1 - x^2 + x^4 - x^6 + x^8.........$

Step 2: Find the derivative of y, i.e. y'

$y' = -2x + 4x^3 - 6x^5 + 8x^7 .............$

Step 3: Substitute y and y' into equation (2) below:

$(1+x^2)y' + 2xy = 0\\\\(1+x^2)(-2x + 4x^3 - 6x^5 + 8x^7......) + 2x(1 - x^2 + x^4 - x^6 + x^8.......) = 0\\\\-2x+ 4x^3 - 6x^5 + 8x^7........ - 2x^3 +4x^5 - 6x^7 + 8x^9 ......+ 2x - 2x^3 + 2x^5 - 2x^7 + 2x^9...... = 0\\\\0 = 0$

(Verified)

Since the LHS = RHS = 0, the given power series $y =\sum^{\infty}_{n=0} {(-1)^n x^{2n}}$ is a solution of the differential equation (1+x^2)y' + 2xy = 0

6 0

3 years ago