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Oksi-84 [34.3K]
3 years ago
9

50 POINTS QUESTION ATTACHED. CHOOSE TWO CORRECT ANSWERS!

Mathematics
1 answer:
OlgaM077 [116]3 years ago
4 0

Answer:

B and D

Step-by-step explanation:

A difference of squares has the form a² - b²

4p² - 9 = (2p)² - 3² ← is a difference of squares

q² - 36 = q² - 6² ← is a difference of squares

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36$+4$=40$

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6 0
3 years ago
Pls, can you help me? thx.<br><img src="https://tex.z-dn.net/?f=%20%5Ccos%28x%20%2B%20%5Cfrac%7B%5Cpi%7D%7B3%7D%29%20%5Cgeqslant
Veseljchak [2.6K]

For x between -\pi and \pi, we have

  • \cos x=\frac{\sqrt2}2=\frac1{\sqrt2} when x=\pm\frac\pi4;
  • \cos x=1 for x=0; and
  • \cos x=0 for x=\pm\frac\pi2

\cos x is continuous over its domain, so the intermediate value theorem tells us that

\cos x\ge\frac{\sqrt2}2

is true for -\frac\pi4\le x\le\frac\pi4.

For all x, we take into account that \cos x is 2\pi-periodic, so the above inequality can be expanded to

-\dfrac\pi4\le x+2n\pi\le\dfrac\pi4

where n is any integer. Equivalently,

-\dfrac\pi4-2n\pi\le x\le\dfrac\pi4-2n\pi

To get the corresponding solution set for

\cos\left(x+\dfrac\pi3\right)\ge\dfrac{\sqrt2}2

simply replace x with x+\frac\pi3:

-\dfrac\pi4-2n\pi\le x+\dfrac\pi3\le\dfrac\pi4-2n\pi

\implies\boxed{-\dfrac{7\pi}{12}-2n\pi\le x\le-\dfrac\pi{12}-2n\pi}

7 0
3 years ago
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