Adding and subtracting polynomials

Mathematics

2 answers:

Colt1911 [192]2 years ago

7 0

Answer:

2x-6

Step-by-step explanation:

Ghella [55]2 years ago

3 0

Your answer is going to be:

2x-6

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PLEASE HELPPPP ASAP!!!! I REALLY NEED IT! WILL MARK BRAINLIEST!!!!

11Alexandr11 [23.1K]

Answer:

3.3

Step-by-step explanation:

d = (10 - 4) + (8 - 3)

d = 6 + 5

d = square root of 11

d = 3.31662479 or 3.3

I'll be expecting that brainliest :)

6 0

3 years ago

20 POINTS!!!

notka56 [123]

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

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