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Svetach [21]
3 years ago
5

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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it looks as though you made 72 of your goal so far for the month you only have 5 days left to reach 100% of your goal let me rem
dezoksy [38]
72% can also be written as 0.72, so take 0.72 x 45 = 32.4     

Subtract 32.4 from 45    45 - 32.4 = 12.6    You need 12.6 renewals in 5 days.
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3 years ago
Read 2 more answers
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}

                                        = \frac{2}{5}

Therefore, average rate of change in the given interval is \frac{2}{5} .

7 0
3 years ago
If f(5) and the equation is f(x)=x to the2nd power +2x
scoundrel [369]
The answer would be 35
5 0
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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2 years ago
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BaLLatris [955]
20.9984 rounded = 21
6 0
2 years ago
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