$9x^6 + 6x^3y + y^2 \\ \\ = {(3 {x}^{3} )}^{2} + 2.(3x^3).y +( y)^2 \\ \\ = \bigg({3 {x}^{3} } + y \bigg)^{2} \\ \\ = \bigg({3 {x}^{3} } + y \bigg) \bigg({3 {x}^{3} } + y \bigg) \\ \\ \\ \\$

4 0

2 years ago

Add and simplify 1/5 + 1/10

aliya0001 [1]

In order to add these fractions we have to make a common denominator.

1/5 can turn into 2/10 Its the same thing. And its easier to solve with.

now lets multiply.

$\frac{2}{10}$ × $\frac{1}{10}$

Now we add across

2 + 1 = 3

------------

10 + 10 = 10 ( When adding fractions like these the denominator stays the same)

So your answer is...

$\frac{3}{10}$

Good Luck! :)

7 0

3 years ago

To solve the inequality 9x >-4, the inequality sign must be reversed.

Anna35 [415]

Answer:

false

Step-by-step explanation:

you only need to reverse the sign when you are dividing by a negative number

7 0

3 years ago

Read 2 more answers

3x + 6y = 14<br>2x + 8y = 2

solmaris [256]

Step-by-step explanation:

x=14-6y/3

substituting x=14-6y/3 into eqn 2

2(14-6y/3)+8y=2

multiplying through by 3

2(14-6y/3)×3 +8y×3=2×3

2(14-6y)+24y=6

28-12y+24y=6

28-12y=6

28-6=12y

22=12y

22/12=y

finding x

2x+8(22/12)=2

2x+44/3=2

2x=2-44/3

x=(-38/3)/2

×=-38/6

7 0

2 years ago