I really need help with this. Please and thank you.

Mathematics

1 answer:

trasher [3.6K]2 years ago

5 0

Answer:

Step-by-step explanation:

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Who knows how to do this ughhh I need help!

suter [353]

Answer:

$\large\boxed{\sqrt{56x^{17}}=2x^8\sqrt{14x}}$

Step-by-step explanation:

$Domain:\ x\geq0\\\\\sqrt{56x^{17}}=\sqrt{4\cdot14\cdot x^{16+1}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{4\cdot14\cdot x^{16}\cdot x^1}\\\\\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt4\cdot\sqrt{14}\cdot\sqrt{x^{16}}\cdot\sqrt{x}=2\cdot\sqrt{14}\cdot\sqrt{x^{8\cdot2}}\cdot\sqrt{x}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=2\cdot\sqrt{14}\cdot\sqrt{(x^8)^2}\cdot\sqrt{x}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2\cdot\sqrt{14}\cdot x^8\cdot\sqrt{x}=2x^8\sqrt{14x}$

8 0

3 years ago

PLEASE HELP ME FAST! THIS IS A SIMPLE TRIG PROBLEM BUT I SKIPPED CLASS.

NikAS [45]

we can factor the whole thing:

(2sin(x) -1)(sin(x)+1) = 0.

Therefore, sin(x) = -1 and sin(x) = 1/2.

For the first one x = 3π/2 and the second is π/6 and 5π/6

So 3π/2, π/6 and 5π/6 are the solutions.

I do kind of have a problem with this because it doesn't mention if you should go over 360°. Otherwise, you have to add in an 2nπ into the equations like 3π/2 + 2nπ; $n \in \mathbb{W}$