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il63 [147K]
3 years ago
9

What is -6/-3 simplified

Mathematics
1 answer:
Setler [38]3 years ago
3 0

Answer:

2

Step-by-step explanation:

3 goes into 6, 2 times and negative on negative is positive

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Solve for q. Write your answer as “q=___”
wolverine [178]

Answer:

7q-46= 3q +6

7q-3q = 6+ 46

4q= 52

4q/4= 52/4

q= 13

3 0
3 years ago
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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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It’s the letter “B” I think
8 0
3 years ago
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The hypotenuse of a right triangle is four times the length of one of the legs. The length of the other leg is 1500⎯⎯⎯⎯⎯⎯⎯⎯√ fee
Mrac [35]

Answer:

Step-by-step explanation:

So this is just playing with the pythagorean theorem

a^2 + b^2 = c^2 with a and b as legs and c as the hypotenuse.

Let's say a is equal to 1500 so then c is 4 times b

1500^2 + b^2 = (4b)^2  Now we just treat this like a normal algebraic expression

2,250,000 = 16b^2 - b^2

2,250,000 = 15b^2

150,000 = b^2

sqrt(150,000) = b

100sqrt(15) = b

You can check this now.

1500^2 + sqrt(150,000)^2 = c^2

sqrt(2,250,000 + 150,000) = c

sqrt(2,400,000) = c

400sqrt(15) = c Which fits the requirement of c = 4b

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3 years ago
Write the following numbers in order from least to greatest,
PIT_PIT [208]

Answer: D

Step-by-step explanation:

Look at the exponents first and order them from least to greatest

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3 years ago
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