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Harrizon [31]
3 years ago
9

Drag each label to the correct location on the table. Each label can be used more than once, but not all labels will be used.

Mathematics
1 answer:
vichka [17]3 years ago
8 0

Answer:

Written in explanation

Step-by-step explanation:

Polynomial 1:

Simplified form:

(x-\frac{1}{2} )(6x +2) = 6x^2 -3x + 2x - 1 = 6x^2 -x -1

Name by number of terms:

There are 3 terms, so this is a trinomial.

Polynomial 2:

Name by degree: The highest power of x is 1, so this is linear.

Name by number of terms: There are two terms, so this is binomial.

Polynomial 3:

Simplified Form:

4(5x^2-9x+7) +2(-10x^2+18x-13) = \\20x^2 - 36x +28 -20x^2 + 26x-26 =\\2

Name by degree: There are no x terms, so this is a constant.

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X1= -1, X2= 2

Step-by-step explanation:

find the x-intercept/zero

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12 divided 934.8 and have fun
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What are the correct answers to this geometry question?
Maru [420]

Answer:

Ti = 5

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Step-by-step explanation:

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The two lines represent the amount of water filling over time in two tanks of the same size.
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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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