Help with math please!!

Factor completely

luda_lava [24]

Answer:

B)

D)

Step-by-step explanation:

1. $7x^3y+14x^2y^3-7x^2y^2$

The GCF of all the term of the above polynomial is $7x^2y$ , hence we take it outside and form a bracket

$7x^2y(x+2y^2-y)$

The polynomial within the bracket can not be factorized further hence this is our final answer. Option (B) is the right answer

2. $5x(x+3)+6(x+3)$

The GCF of all the term of the above polynomial is $(x+3)$ , hence we take it outside and form a bracket

$(x+3)(5x+6)$

The polynomial within the bracket can not be factorized further hence this is our final answer. Option (B) is the right answer

3. $8x^5+2x^4+4x^2$

The GCF of all the term of the above polynomial is $2x^2$ , hence we take it outside and form a bracket

$2x^2(4x^3+x^2+2)$

The polynomial within the bracket can not be factorized further hence this is our final answer. Option (D) is the right answer

4 0

3 years ago

Jessica used Fraction 2 over 3 yard of fabric to make a scarf. Can she make 2 of these scarves with Fraction 1 and 3 over 4 yard

aleksklad [387]

Use Socratic it will give you the right answers

6 0

3 years ago

New question, what is 100 + 100 + 100 x 30?<br><br> More points!

vovikov84 [41]

Answer:

9,000

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Simplify: 6a - 9+<br> 10a.

stepan [7]

16a - 9 I think that’s the answer hope this helped

7 0

3 years ago

37. Socks. In your sock drawer you have 4 blue socks, 5 grey socks, and 3 black ones. Half asleep one morning, you grab 2 socks

mihalych1998 [28]

Answer:

a) $\frac{1}{22}$

b) $\frac{7}{22}$

c) $\frac{5}{11}$

d) $\frac{35}{66}$

Step-by-step explanation:

Given,

Number of blue socks = 4,

Grey socks = 5,

Black socks = 3,

Total socks = 4 + 5 + 3 = 12,

Ways of choosing any 2 socks = $^{12}C_2$

$=\frac{12!}{2!10!}$

$=\frac{12\times 11}{2}$

$=6\times 11$

= 66,

a) Ways of choosing 2 blue socks = $^3C_2$

= 3,

Since, $\text{Probability}=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

Thus, the probability of selecting 2 blue socks = $\frac{3}{66}=\frac{1}{22}$

b) Ways of choosing 2 shocks, non of them are grey socks = $^7C_2$

$=\frac{7!}{2!5!}$

$=7\times 3$

$=21$

Thus, the probability of selecting no grey socks = $\frac{21}{66}$

$=\frac{7}{22}$

c) ways of selecting atleast 1 black sock = ways of selecting 1 black sock + ways of selecting 2 black socks

$=^3C_1\times ^9C_1+^3C_2$

$=3\times 9 + 3$

$=27 + 3$

= 30,

Thus, the probability of selecting at least 1 black sock = $\frac{30}{66}$

$=\frac{5}{11}$

d) Ways of selecting a green sock = $^5C_1\times ^7C_1$

= 35,

Thus, the probability of selecting a green sock = $\frac{35}{66}$

3 0

3 years ago