Simplify.

1 answer:

3 0

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^m)^n=a^{n\cdot m}\\---------------------\\\\\left(4a^5b^6\right)^4=4^4(a^5)^4(b^6)^4=256a^{5\cdot4}b^{6\cdot4}=256a^{20}b^{24}$

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I really need help with this ?

Digiron [165]

The answer is g. 47.1

you use the formula
c= π d
then plug is 15 for d and you should get the answer

8 0

3 years ago

What times 4/9 = -8?

djverab [1.8K]

Answ

This is false

Step-by-step explanation:

3 0

3 years ago

20 POINTS!!!

notka56 [123]

$\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------$

$(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s$