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3 years ago

Simplify y 2 + 11y - 6y + y 2. y4 + 5 y4 + 5y 2y 2 + 5 2y 2 + 5y

Mathematics

2 answers:

kolezko [41]3 years ago

5 0

Answer:

$\Large \boxed{2y^2 + 5y}$

Step-by-step explanation:

$\large y^2 + 11y - 6y + y^2$

$\sf Group \ the \ like \ terms.$

$\large (y^2 +y^2) + (11y - 6y)$

$\sf Combine \ all \ like \ terms.$

$\large 2y^2 + 5y$

Dmitrij [34]3 years ago

3 0

Answer:

2y^2 +5y

Step-by-step explanation:

y^2 + 11y - 6y + y^2

Combine like terms

2y^2 +5y

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Given the function defined in the table below, find the average rate of

AURORKA [14]

Answer:

Average rate of change = $\frac{2}{5}$

Step-by-step explanation:

Average rate of change of a function 'f' in the interval a ≤ x ≤ b is given by,

Average rate of change = $\frac{f(b)-f(a)}{b-a}$

We have to find the average rate of change in the interval 20 ≤ x ≤ 65

From the table attached,

f(65) = 32

f(20) = 14

Average rate of change = $\frac{32-14}{65-20}$

= $\frac{18}{45}$

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scoundrel [369]

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3 years ago

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

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