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anastassius [24]
3 years ago
5

Can someone please prove that the expression (x-y)+(y-z)+(z-x) is equivalent to 0

Mathematics
1 answer:
ELEN [110]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

(x-y)+(y-z)+(z-x)\\\\=x-y+y-z+z-x\\\\=(x-x)+(-y+y)+(-z+z)\\\\=0+0+0\\\\=0

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Given the functions a(x) = 3x − 12 and b(x) = x − 9, solve a[b(x)].
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A(x) = 3x - 12
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a(b(x)) = a(x - 9) = 3(x - 9) - 12 = 3x - 27 - 12 = 3x - 39
a(b(x)) = 3x - 39
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3 years ago
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Hoochie [10]

Answer:

\large\boxed{\sin2\theta=\dfrac{\sqrt3}{2},\ \cos2\theta=\dfrac{1}{2}}

Step-by-step explanation:

We know:

\sin2\theta=2\sin\theta\cos\theta\\\\\cos2\theta=\cos^2\theta-\sin^2\thet

We have

\sin\theta=\dfrac{1}{2}

Use \sin^2\theta+\cos^2\theta=1

\left(\dfrac{1}{2}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{4}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{1}{4}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{4}{4}-\dfrac{1}{4}\\\\\cos^2\theta=\dfrac{3}{4}\to\cos\theta=\pm\sqrt{\dfrac{3}{4}}\to\cos\theta=\pm\dfrac{\sqrt3}{\sqrt4}\to\cos\theta=\pm\dfrac{\sqrt3}{2}\\\\\theta\in[0^o,\ 90^o],\ \text{therefore all functions have positive values or equal 0.}\\\\\cos\theta=\dfrac{\sqrt3}{2}

\sin2\theta=2\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\sqrt3}{2}\\\\\cos2\theta=\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2}

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3 years ago
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Answer:

LN = 5x + 9

Step-by-step explanation:

LN = x + 9 + 4x

LN = 5x + 9

4 0
3 years ago
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