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3 years ago

PLS GIVE ANSWER AND TELL ME EXACTLY HOW YOU GOT IT

Mathematics

1 answer:

PIT_PIT [208]3 years ago

8 0

A) Y U Z { B, C, D, F, G, H, K, T, L, M, N, W}
b) n(YUZ) {5} (n is the number of elements in the set)
c) X AND Y {K}
d) n(X AND Y) {16}

I'm not sure of that notation for e and f. Can you tell me what the X\Y means?

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Factor out the GCF. 12ax3 +20bx² + 32cx [?]x([]ax +[ ]bx + [ ]c) Enter

olganol [36]

Answer:

4x(3ax²+5bx+8c)

Step-by-step explanation:

4 is the highest number that fits into all terms of the expression :

So outside the bracket will be 4x :

To find what is inside the bracket we divide each term by 4x

12ax³÷4x = 3ax²

20bx²÷4x=5bx

32cx÷4x=8c

So our final answer is :

4x(3ax²+5bx+8c)

Hope this helped and have a good day

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2 years ago

Jeremey performs the same operations on four values of x. He records each resulting y-value in a table, as shown below. Which eq

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2 years ago

-k+3k combine like terms to create an equivalent expression

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Step-by-step explanation:

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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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Answer:

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