What is the values of the soul out ion to the inequality 9 < 3x + 4?

Mathematics

1 answer:

HACTEHA [7]3 years ago

5 0

Answer:

x>5/3

Step-by-step explanation:

9<3x+4

3x+4>9

3x>9-4

3x>5

x>5/3

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For f(x)=2x+1 and g(x)=x^2-7, find (f+g)(x) A. 2x^2-15 B.X^2+2x-6 C.2x^3-6 D.x^2+2x+8

Advocard [28]

Answer:

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) , thus

f(x) + g(x)

= 2x + 1 + x² - 7 ← collect like terms

= x² + 2x - 6 → C

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3 years ago

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Alla [95]

Answer:

x=18, 96-24=72, x+3x=72, 4x=72, x=18

Step-by-step explanation:

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3 years ago

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Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$