Completing the squares tells us exactly what square root we will need to calculate our answer.

Using the formula below you can find these answers:
$A(x-h)^2 + K = 0
(x-5)^ 2 = - K/A
$

Does this make sense?

4 0

3 years ago

Verify the trigonometric identities

snow_lady [41]

1)

here, we do the left-hand-side

$\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2
\\\\\\\
[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]
\\\\\\
2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2$

2)

here we also do the left-hand-side

$\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)
\\\\\\
\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}
\\\\\\
\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)$

3)

here, we do the right-hand-side

$\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}
\\\\\\
\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}
\\\\\\
\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}$

8 0

3 years ago