3 years ago

Simplify completely 12x^3-4x^2+8x over -2x

Mathematics

1 answer:

sergejj [24]3 years ago

4 0

Hello :
<span> (12x^3-4x^2+8x) / -2x = - 6x²+2x - 4</span>

You might be interested in

Gabrielle is 12 years younger than Mikhail. The sum of their ages is 36 . What is Mikhail's age?

ololo11 [35]

X = age of Mikhail

x+x-12 = 36
2x= 36 + 12
2x = 48
x = 48/2
x= 24

3 0

3 years ago

Read 2 more answers

There were 48 blue, 48 green and 48 yellow crayons. Also there were 72 red crayons and 120 colored pictures. What is the largest

Elis [28]

48 sets i believe since there are only 48 maximum of blue, green and yellow crayons therefore the 49th set would only contain a red crayons and a picture

3 0

2 years ago

Read 2 more answers

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$