But that doesn't really make sense, because if x has already been solved, then the equation makes sense, because if we substitute x into it, then we have 82-24, which is 58. I could be wrong... sooo... but I hope it helps?

5 0

3 years ago

What is the midpoint of (-1,-6), (-6, 5)

kirill115 [55]

<h3>Answer:</h3><h3>-3.5,-0.5</h3>

Step-by-step explanation:

6 0

4 years ago

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function? (9x + 7)(4x + 1)(3x + 4) = 0

DanielleElmas [232]

Answer:

Step-by-step explanation:

3 roots

namely -7/9,-1/4.-4/3

4 0

3 years ago

Multiply.

Nady [450]

$\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}$

<h2>Explanation:</h2>

Here we have the following expression:

$3\sqrt{22}\sqrt{58}\sqrt{18}$

So we need to simplify that radical expression. By property of radicals we know that:

$\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}$

So:

$3\sqrt{22}\sqrt{58}\sqrt{18}=3\sqrt{22\times 58 \times 18}=3\sqrt{22968}$

The prime factorization of 22968 is:

$22968=2^3\cdot 3^2\cdot11\cdot 29$

Hence:

$3\sqrt{22968}=3\sqrt{2^3\cdot 3^2\cdot11\cdot 29}=3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29}$

By property:

$\sqrt[n]{a^n}=a$

So:

$3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29} \\ \\ 3(2)(3)\sqrt{2\cdot 11\cdot 29}=18\sqrt{638}$

Finally:

$\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}$

<h2>Learn more:</h2>

Radical expressions: brainly.com/question/13452541

#LearnWithBrainly

3 0

3 years ago