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3 years ago

What is the domain and range for: y = arcsin(tanx)

Mathematics

1 answer:

Semmy [17]3 years ago

7 0

The graph looks like this, on the enclosed pic:
One feature is that it's periodic and torn (has cut-off points), meaning the domain is the same as in case of tan(x): x€R and x =/= π/2+πn.
The range equals the range of arcsin(x): -π/2<=y<=π/2 OR y€[-π/2;π/2]
Hope could understand and if it helped! :)

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Answer:

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3 years ago

A 10 2/3 inch long pipe is cut into 8 equal length pieces, how long is each piece

storchak [24]

Answer:

$\frac{4}{3}$ inches

Step-by-step explanation:

The total length of pipe is 10 2/3rd inches

It is cut into 8 equal pieces

We need to divide total length by 8 to get length of each piece. First, we have to convert mixed number (total length) to improper fraction by using formula shown below:

$a\frac{b}{c}=\frac{(a*c)+b}{c}$

So,

$10\frac{2}{3}=\frac{32}{3}$

Each piece length is:

$\frac{\frac{32}{3}}{8}\\=\frac{32}{3}*\frac{1}{8}\\=\frac{32}{24}\\=\frac{4}{3}$

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3 years ago

Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of 1.1 years. Suppose also that exactl

joja [24]

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3 years ago

PLEASEEE HELP ! I WOULD APPRECIATE IT VERY MUCH :)

azamat

hi!

so the original equation is (x^2 -121) / x+ 11.

Those two equations look similar...

well, we know that the top equation looks like a^2 - b^2, and that equation iis equivalent to this equation: (a+b)(a-b).

So if we factor that out, we get:

( (x + 11) * (x-11) )/ (x + 11)

we can cancel the x+11 on the top and the x+11 on the bottom out.

that leaves us with x-11.

Hope this helped!

8 0

3 years ago

Suppose you pick two cards from a deck of 52 playing cards. What is the probability that they are both queens?

photoshop1234 [79]

Answer:

0.45% probability that they are both queens.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes

The combinations formula is important in this problem:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes

You want 2 queens. Four cards are queens. I am going to call then A,B,C,D. A and B is the same outcome as B and A. That is, the order is not important, so this is why we use the combinations formula.

The number of desired outcomes is a combinations of 2 cards from a set of 4(queens). So

$D = C_{4,2} = \frac{4!}{2!(4-2)!} = 6$

Total outcomes

Combinations of 2 from a set of 52(number of playing cards). So

$T = C_{52,2} = \frac{52!}{2!(52-2)!} = 1326$

What is the probability that they are both queens?

$P = \frac{D}{T} = \frac{6}{1326} = 0.0045$

0.45% probability that they are both queens.

4 0

3 years ago