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3 years ago

Nine is less than a number is no more than 8 and no less than 3

Mathematics

2 answers:

ArbitrLikvidat [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Let <em>n - </em><em>number</em>

Nine less than a number: <em>n - 9</em>

is no more than 8: <em>≤ 8</em>

and no less than 3: ≥<em> 3</em>.

We have the inequality:

$3\leq n-9\leq8\qquad\text{add 9 to all sides}\\\\3+9\leq n-9+9\leq8+9\\\\12\leq n\leq17$

9966 [12]3 years ago

5 0

Answer:

3 <= x <= 8

Step-by-step explanation:

A number is denoted as an arbitrary variable, let's call it x.

We have a double inequality:

1. No more than 8 means less than or equal to 8

2. No less than 3 means greater than or equal to 3

3 <= x <= 8

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The value of the expression $2\cos(t) + \tan^2(t)$ is 4 and the exact value of $\sin^{-1}(\sin(\frac{5\pi}{3}))$ is $\frac{5\pi}{3}$

<h3>How to determine the trigonometry expression?</h3>

The point on the unit circle is given as:

$P = (\frac 12, \frac{\sqrt{3}}2)$

A point on a unit circle is represented as: (x,y), such that:

cos(t) = x and sin(t) = y.

This means that:

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Calculate tan(t) using:

$\tan(t) = \frac{\sin(t)}{\cos(t)}$

So, we have:

$\tan(t) = \frac{\frac{\sqrt 3}{2}}{1/2}$

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$\tan(t) = \sqrt 3$

The expression is then calculated as:

$2\cos(t) + \tan^2(t) = 2 * \frac12 + (\sqrt 3)^2$

Evaluate each term

$2\cos(t) + \tan^2(t) = 1 + 3$

Evaluate the sum

$2\cos(t) + \tan^2(t) = 4$

Hence, the value of the expression $2\cos(t) + \tan^2(t)$ is 4

<h3>How to solve the arcsin expression?</h3>

The expression is given as:

$\sin^{-1}(\sin(\frac{5\pi}{3}))$

As a general rule, the arc sine of sine x is x.

This means that:

$\sin^{-1}(\sin(\frac{5\pi}{3})) = \frac{5\pi}{3}$

Hence, the exact value of $\sin^{-1}(\sin(\frac{5\pi}{3}))$ is $\frac{5\pi}{3}$

Read more about trigonometry expressions at:

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