1. y = 7
2. 0
3. m = -1
4. k = -4
5. x = -6
6. There is no solution for these equations.

Please let me know if you want explanations/work shown!! Hope this helps :)

7 0

4 years ago

Coefficiants of (2x+y)^4

sattari [20]

By the binomial theorem,

$(2x+y)^4=\displaystyle\sum_{k=0}^4\binom 4k(2x)^{4-k}y^k=\sum_{k=0}^4\binom 4k2^{4-k}x^{4-k}y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

Then the coefficients of the $x^{4-k}y^k$ terms in the expansion are, in order from $k=0$ to $k=4$ ,

$\dbinom 402^{4-0}=1\cdot2^4=16$

$\dbinom412^{4-1}=4\cdot2^3=32$

$\dbinom422^{4-2}=6\cdot2^2=24$

$\dbinom432^{4-3}=4\cdot2^1=8$

$\dbinom442^{4-4}=1\cdot2^0=1$

3 0

3 years ago

Directions: Simplify the following expressions using the imaginary number.

drek231 [11]

Answer:

1 i(sqrt5)

2 7i

3 -9i

4 -64i

5 -5i(sqrt2)

6 -4i(sqrt3)

7 10i(sqrt3)

8 2i(sqrt2)

9 3i(sqrt7)

10 -30i(sqrt2)

Step-by-step explanation:

i is the square root of (-1). We can simplify each one as such.

5 0

2 years ago