What is the answer to this question? Please help!

Mathematics

2 answers:

maks197457 [2]3 years ago

6 0

Answer:

Step-by-step explanation:

$-3\sqrt{2}+3\sqrt{8}= \\\\-3\sqrt{2}+3\sqrt{2\cdot 2^2}= \\\\-3\sqrt{2}+6\sqrt{2}= \\\\3\sqrt{2}$

Hope this helps!

mario62 [17]3 years ago

5 0

3 square root of 2 is the answer

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List 2 integers that are congruent to 4 modulo 14.

vivado [14]

Answer:

32, 46

Step-by-step explanation:

Remember, a is congruent to b modulo d if d divides a-b.

Now, the problem says that b=4 and d=14.

Let a=32. Observe that a-b=28 and $\frac{28}{14}=2$ , then 32 is congruent to 4 modulo 14.

Let a=46. Observe that a-b=46-4=42 and $\frac{42}{14}=3$ , then 46 is congruent to 4 modulo 14.

7 0

3 years ago

Number 2 I need help on how to solve it

Hunter-Best [27]

Let's simplify the following expression:

$\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}$

We get,

$\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}$ $\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}$